How Do You Solve for Landing Speed?
$begingroup$
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
asked Mar 23 '15 at 16:18
RobertRobert
146129
$endgroup$
add a comment |
$begingroup$
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
asked Mar 23 '15 at 16:18
RobertRobert
146129
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Mar 23 '15 at 16:18
RobertRobert
146129
$endgroup$
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
asked Mar 23 '15 at 16:18
RobertRobert
146129
asked Mar 23 '15 at 16:18
RobertRobert
146129
asked Mar 23 '15 at 16:18
RobertRobert
146129
asked Mar 23 '15 at 16:18
RobertRobert
146129
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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active
oldest
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votes
$begingroup$
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:
$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
$endgroup$
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:
$$v = sqrt{50^2 + (-80)^2 + (-20)^2} ft/s = sqrt{9300} ft/s.$$
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
edited Mar 23 '15 at 16:38
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
answered Mar 23 '15 at 16:31
JohnJohn
22.7k32450
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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