A ball of mass 0.4 kg, initially at rest, is kicked directly toward a fence from a point 20 m away, as shown below.
The velocity of the ball as it leaves the kicker’s foot is 16 m/s at angle of 46◦ above the horizontal. The top of the fence is 4 m high. The ball hits nothing while in ﬂight and air resistance is negligible. The acceleration due to gravity is 9.8 m/s^2.
Determine the time it takes for the ball to reach the plane of the fence.

How far above the top of fence will the ball pass? Consider the diameter of the ball to be negligible

What is the vertical component of the velocity when the ball reaches the plane of the fence? Answer in units of m/s

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1. The horizontal velocity component, which does not change, is
Vx = Vo cos46 = 11.11 m/s. Divide the distance to the fence by Vx to obtain the flight time to the plane of the fence.

Use that time (t) and the equation for height above the ground (given below), to find out how far whether, or by how much, the ball clears the fence.

y = (16 sin46)t - (g/2)*t^2

For the vertical component of V,

Vy = 10 sin46 - gt

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2. That is not right drwls

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3. The third question is wrong...

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