How Far Beyond the Fence Does the Ball Land?

[shane99a]

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.

Here's how i did it and its probably wrong so i need help:
V_i = 0
θ = 45°
V_f = ?

0.5mv² = 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_yt - 0.5at²
0 = 6.26t - 4.9t²
t = 1.28s

D = V_xt
D = (6.26)(1.28) = 11 m
Therefore the answer is (11 - 5) 6 m which is the distance the ball traveled 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.

 

[nickjer]

Your initial velocity is not 0 m/s since he originally threw the ball. Also, I wouldn't recommend using conservation of energy for this problem. I recommend using the ballistic equations. For example, there is no reason that your final velocity will be at a 45 degree angle with the horizontal.

 

[rl.bhat]

Hi shane99a, welcome to PF.
Your calculation of v is wrong. It is not the velocity of the projection of the ball.
It is true only if 4 m is the maximum height. But in the problem it is not mentioned.
So you have to use the equation
Dy = vy*t - 1/2*g*t^2...(1)
Substitute t = x/vx and solve for vx. In the given problem x is given.
To find maximum range, put Dy = 0 in eq.(1) and solve for x.