How Do You Solve for Landing Speed?












1












$begingroup$


A ball is thrown eastward into the air from the origin (in the direction of the positive x − axis
). The initial velocity is 50i + 80k. The spin of the ball results in a southward acceleration of 4 ft / s^2 , so the acceleration vector is a -4j - 32k . Where does the ball land and with what
speed?



I found the answer to first part of the question, which is <250, -50, 0> (I am sure this is correct), but I don't understand how to arrive at the second part.



So far, I have calculated
t = 5 seconds
v_xf = 50
v_yf = 20
v_zf = 80



and arrive at the speed sqrt 9300, but the answer is 10 sqrt 33






physics







share|cite|improve this question









$endgroup$












  • $begingroup$
    You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10sqrt{33}$ is incorrect.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:36










  • $begingroup$
    Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:39










  • $begingroup$
    Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks.
    $endgroup$
    – Robert
    Mar 23 '15 at 16:43
















1












$begingroup$


A ball is thrown eastward into the air from the origin (in the direction of the positive x − axis
). The initial velocity is 50i + 80k. The spin of the ball results in a southward acceleration of 4 ft / s^2 , so the acceleration vector is a -4j - 32k . Where does the ball land and with what
speed?



I found the answer to first part of the question, which is <250, -50, 0> (I am sure this is correct), but I don't understand how to arrive at the second part.



So far, I have calculated
t = 5 seconds
v_xf = 50
v_yf = 20
v_zf = 80



and arrive at the speed sqrt 9300, but the answer is 10 sqrt 33






physics







share|cite|improve this question









$endgroup$












  • $begingroup$
    You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10sqrt{33}$ is incorrect.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:36










  • $begingroup$
    Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:39










  • $begingroup$
    Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks.
    $endgroup$
    – Robert
    Mar 23 '15 at 16:43














1












1








1





$begingroup$


A ball is thrown eastward into the air from the origin (in the direction of the positive x − axis
). The initial velocity is 50i + 80k. The spin of the ball results in a southward acceleration of 4 ft / s^2 , so the acceleration vector is a -4j - 32k . Where does the ball land and with what
speed?



I found the answer to first part of the question, which is <250, -50, 0> (I am sure this is correct), but I don't understand how to arrive at the second part.



So far, I have calculated
t = 5 seconds
v_xf = 50
v_yf = 20
v_zf = 80



and arrive at the speed sqrt 9300, but the answer is 10 sqrt 33






physics







share|cite|improve this question









$endgroup$




A ball is thrown eastward into the air from the origin (in the direction of the positive x − axis
). The initial velocity is 50i + 80k. The spin of the ball results in a southward acceleration of 4 ft / s^2 , so the acceleration vector is a -4j - 32k . Where does the ball land and with what
speed?



I found the answer to first part of the question, which is <250, -50, 0> (I am sure this is correct), but I don't understand how to arrive at the second part.



So far, I have calculated
t = 5 seconds
v_xf = 50
v_yf = 20
v_zf = 80



and arrive at the speed sqrt 9300, but the answer is 10 sqrt 33






physics




physics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 23 '15 at 16:18









RobertRobert

146129




146129












  • $begingroup$
    You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10sqrt{33}$ is incorrect.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:36










  • $begingroup$
    Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:39










  • $begingroup$
    Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks.
    $endgroup$
    – Robert
    Mar 23 '15 at 16:43


















  • $begingroup$
    You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10sqrt{33}$ is incorrect.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:36










  • $begingroup$
    Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form.
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:39










  • $begingroup$
    Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks.
    $endgroup$
    – Robert
    Mar 23 '15 at 16:43
















$begingroup$
You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10sqrt{33}$ is incorrect.
$endgroup$
– rnrstopstraffic
Mar 23 '15 at 16:36




$begingroup$
You're derivatives aren't quite right; their magnitudes are right, but their signs are wrong. It looks like the answer $10sqrt{33}$ is incorrect.
$endgroup$
– rnrstopstraffic
Mar 23 '15 at 16:36












$begingroup$
Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form.
$endgroup$
– rnrstopstraffic
Mar 23 '15 at 16:39




$begingroup$
Also, have you written this out using functions? @john has the right idea that since the unit vectors are orthogonal, we can treat them independently. Thus, the best approach would be to build your velocity function by integrating the acceleration function (using your initial velocity). The position function can be found by integrating the velocity function (using your initial position). If done this way, you've gained all of the information in a usable form.
$endgroup$
– rnrstopstraffic
Mar 23 '15 at 16:39












$begingroup$
Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks.
$endgroup$
– Robert
Mar 23 '15 at 16:43




$begingroup$
Oh okay, yeah I was thinking it was probably the answer that was wrong. No, I should try that now that I think of it. Thanks.
$endgroup$
– Robert
Mar 23 '15 at 16:43










1 Answer
1






active

oldest

votes


















0












$begingroup$

Treat each coordinate independently (since the coordinates are orthogonal and there's no coupling).



After $5$ seconds (when the ball lands), the $x$ component of the velocity will be the same: $50 ft/s$. The $z$ component will be the negative of its starting velocity: $-80 ft/s$. The $y$ component will be its acceleration over that time, times the time: $-4 cdot 5 = -20 ft/s$.



Then, taking the square root of the sum of squares:



$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
    $endgroup$
    – Robert
    Mar 23 '15 at 16:34










  • $begingroup$
    $sqrt{9300}neq 10sqrt{33}$
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:34






  • 1




    $begingroup$
    @rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
    $endgroup$
    – John
    Mar 23 '15 at 16:41











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1 Answer
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$begingroup$

Treat each coordinate independently (since the coordinates are orthogonal and there's no coupling).



After $5$ seconds (when the ball lands), the $x$ component of the velocity will be the same: $50 ft/s$. The $z$ component will be the negative of its starting velocity: $-80 ft/s$. The $y$ component will be its acceleration over that time, times the time: $-4 cdot 5 = -20 ft/s$.



Then, taking the square root of the sum of squares:



$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
    $endgroup$
    – Robert
    Mar 23 '15 at 16:34










  • $begingroup$
    $sqrt{9300}neq 10sqrt{33}$
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:34






  • 1




    $begingroup$
    @rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
    $endgroup$
    – John
    Mar 23 '15 at 16:41
















0












$begingroup$

Treat each coordinate independently (since the coordinates are orthogonal and there's no coupling).



After $5$ seconds (when the ball lands), the $x$ component of the velocity will be the same: $50 ft/s$. The $z$ component will be the negative of its starting velocity: $-80 ft/s$. The $y$ component will be its acceleration over that time, times the time: $-4 cdot 5 = -20 ft/s$.



Then, taking the square root of the sum of squares:



$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
    $endgroup$
    – Robert
    Mar 23 '15 at 16:34










  • $begingroup$
    $sqrt{9300}neq 10sqrt{33}$
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:34






  • 1




    $begingroup$
    @rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
    $endgroup$
    – John
    Mar 23 '15 at 16:41














0












0








0





$begingroup$

Treat each coordinate independently (since the coordinates are orthogonal and there's no coupling).



After $5$ seconds (when the ball lands), the $x$ component of the velocity will be the same: $50 ft/s$. The $z$ component will be the negative of its starting velocity: $-80 ft/s$. The $y$ component will be its acceleration over that time, times the time: $-4 cdot 5 = -20 ft/s$.



Then, taking the square root of the sum of squares:



$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$






share|cite|improve this answer











$endgroup$



Treat each coordinate independently (since the coordinates are orthogonal and there's no coupling).



After $5$ seconds (when the ball lands), the $x$ component of the velocity will be the same: $50 ft/s$. The $z$ component will be the negative of its starting velocity: $-80 ft/s$. The $y$ component will be its acceleration over that time, times the time: $-4 cdot 5 = -20 ft/s$.



Then, taking the square root of the sum of squares:



$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 23 '15 at 16:38

























answered Mar 23 '15 at 16:31









JohnJohn

22.7k32450




22.7k32450












  • $begingroup$
    Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
    $endgroup$
    – Robert
    Mar 23 '15 at 16:34










  • $begingroup$
    $sqrt{9300}neq 10sqrt{33}$
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:34






  • 1




    $begingroup$
    @rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
    $endgroup$
    – John
    Mar 23 '15 at 16:41


















  • $begingroup$
    Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
    $endgroup$
    – Robert
    Mar 23 '15 at 16:34










  • $begingroup$
    $sqrt{9300}neq 10sqrt{33}$
    $endgroup$
    – rnrstopstraffic
    Mar 23 '15 at 16:34






  • 1




    $begingroup$
    @rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
    $endgroup$
    – John
    Mar 23 '15 at 16:41
















$begingroup$
Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
$endgroup$
– Robert
Mar 23 '15 at 16:34




$begingroup$
Wait, but sqrt 9300 calculates to 96.43, while 10 sqrt 33 calculates to 57.45
$endgroup$
– Robert
Mar 23 '15 at 16:34












$begingroup$
$sqrt{9300}neq 10sqrt{33}$
$endgroup$
– rnrstopstraffic
Mar 23 '15 at 16:34




$begingroup$
$sqrt{9300}neq 10sqrt{33}$
$endgroup$
– rnrstopstraffic
Mar 23 '15 at 16:34




1




1




$begingroup$
@rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
$endgroup$
– John
Mar 23 '15 at 16:41




$begingroup$
@rnrstopstraffic Not sure where my mind was when I typed that! Fixed.
$endgroup$
– John
Mar 23 '15 at 16:41


















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