An archer shoots an arrow with a velocity of 45.0 m/s at an angle of 50.0 degree

Question

An archer shoots an arrow with a velocity of 45.0 m/s at an angle of 50.0 degrees with the horizontal. An assistant standing on the level ground 150 m down-range from the launch point throws an apple straight up with the minimum initial speed necessary to meet the path of the arrow.
a. What is the initial speed of the apple?
b. At what time after the arrow launch should the apple be thrown so that the arrow hits the apple?

PS: I know the answer is 30.4 m/s, and 2.09 s later. But how to you get these answers

Explanation / Answer

Horizontal velocity = 45 cos 50

So, time taken to reach 150 m is

t = 150/45 cos 50 = 5.19 s

(a)Height of arrow at that time = v sin * t - (1/2)gt^2 = 178.9 - 131.9 = 47 m

Maximum height of vertically thrown object = u^2/2g

47 = u^2 / 19.6

u = 30.4 m/s

(b)time taken by the apple to reach the point = u/g = 30.4 / 9.8 = 3.1 s

So, time after which it has to be thrown = 5.19 - 3.1 = 2.09 s

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