SOLUTION: Hi again. I'm sorry to bother you guys again but I have one more question. Here it is:
A ball is thrown up at 50 m/s at an angle of 37 degrees. How far has the ball traveled (h
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-> SOLUTION: Hi again. I'm sorry to bother you guys again but I have one more question. Here it is:
A ball is thrown up at 50 m/s at an angle of 37 degrees. How far has the ball traveled (h
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Question 810870: Hi again. I'm sorry to bother you guys again but I have one more question. Here it is:
A ball is thrown up at 50 m/s at an angle of 37 degrees. How far has the ball traveled (horizontally) when the ball is 25 meters above the ground?
My AP Physics teacher gave me the answer for this problem, which can be either 39.9 meters OR 205 meters, but she wants me to show how I would get to either of those answers by showing my work. I think the equations I'm using for the problem like v= vi + at to get the time and v = x/t to get the distance are right but the fact that the ball is 25 meters above the ground might be the reason why I can't get either 39.9 m or 205 m. I don't know where 25 meters would fit into any of the equations. Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! angle of projection = 37 deg
calculate the time taken to reach a height of 25 m
The vertical component is v sin
y=25=ut+1/2 (-g)t^2 where t is the time in seconds
25=30t-1/2*9.8t^2
4.9t^2-30t+25=0
t= 5.13 & 1
in the horizontal direction d=rt
d=v cos*t
=v cos 37 *1
=39.9 meters
OR
d=vcos 37 * 5.13
=204.85 meters
