Put v = 19 m/s and alpha = 41°.
v cos(alpha) is the horizontal speed of the ball and v sin(alpha) is the vertical speed of the ball. The equations for the vertical and horizontal motion of the ball are independent of each other. This allows you to consider only the vertical motion to find out when the ball has fallen back to the same height above the ground from which it was thrown.
The initial velocity in the vertical direction is v sin(alpha) and this changes as a function of time according to:
v sin(alpha) - g t
Now, you know that due to conservation of energy when the ball is back at the same heightit will ahve the same speed, this means that the velocity will have canged sign. So, the time it takes for the ball to be bsck at the initial height follows from:
v sin(alpha) - g t = - v sin(alpha)
t folows from this that:
t = 2 v sin(alpha)/g
The horizontal speed of the ball is time independent, there fore the ball will have tavelled a distance of:
d = v cos(alpha) t =
2 v^2 cos(alpha) sin(alpha)/g =
v^2 sin(2 alpha)/g
Gioven this distance d from the point the ball is thrown, you can calculate the distance to the catcher, you know that the catcher has to cover that distance in a time of t.
A ball is thrown at 19 m/s at 41° above the horizontal. Someone located 30 m away along the line3 of the path starts to run just as the ball is thrown. How fast, and in which direction, must the catcher run to catch the ball at the level from which it was thrown?
not quite sure how to go about this..
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