A kid is 5 m from a fence that is 4 m high. He throws a ball at 45° from the horizontal which just grazes the fence. How far beyond the fence does the ball land? You may assume that the ball was thrown from the same level as the ground on the far side of the fence.(adsbygoogle = window.adsbygoogle || []).push({});

Here's how i did it and its probably wrong so i need help:

V_{i}= 0

θ = 45°

V_{f}= ?

0.5mv^{2}= mgh

v = √(2gh)

v = √(19.6*4)

V_{f}= 8.85 m/s

V_{y}= 8.85sin45 = 6.26 m/s = V_{x}

D_{y}= V_{y}t - 0.5at^{2}

0 = 6.26t - 4.9t^{2}

t = 1.28s

D = V_{x}t

D = (6.26)(1.28) = 11 m

Therefore the answer is (11 - 5) 6 m which is the distance the ball travelled beyond the fence.

It looks correct except for the height which I don't know if picking 4 was the right idea and would screw up my entire answer, so please someone help.

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# Homework Help: Sin contest- Projectile motion

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