# a ball of mass 0.2 kg, initially at rest, is kicked directly toward a fence from a point 20m away

a ball of mass 0.2 kg, initially at rest, is kicked directly toward a fence from a point 20m away. The velocity of the ball as it leaves the kicker's foot is 20m/s at angle of 40º above the horizontal ground. The top of the fence is 4m high. THe kicker's foot is in contact with the ball for 0.05s. The ball hits nothing while in flight and air resistance is negligible. The acceleration due to gravity is 9.8m/s^2.

1) Determine the time it takes for the ball to reach the plane of the fence. Answer in units of s.

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

I think the equation: cos(38) x 17m/s = 1/2 a t^2 comes in somewhere in this question, but i don't know how it fits into this question. I'm seriously stuck D: can anyone help me? thanks in advance!

1) Divide the distance to the fence by the horizontal component of the kickoff velocity, which is 20 m/s* cos 40. That will give you the time to reach the fence

2) The vertical component of velocity at time t is

Vy = 20 m/s sin 40 – g t

The equation you were thinking of is incorrect, and not even dimensionally consistent. You cannot have m/s on one side of an equation and meters (1/2 a t^2) on the other.

2) when you say Vy=20m/s sin40 – gt,

is g=9.8 and is t=answer from first question? because when i did the questions, i got a negative answer and you can't have negative velocity can you? thanks a lot!