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.

Question

The velocity of the ball as it leaves the kicker's foot is 17 m/s at angle of 35 degree above the horizontal. The top of the fence is 3 m high. The ball hits nothing while in flight and air resistance is negligible.
The acceleration due to gravity is 9.8 m/s^
Determine the time it takes for the ball to reach the plane of the fence.
Answer in units of s.

Answer 1

How far above the top of fence will the ball pass? Consider the diameter of the ball to be negligible.
Answer in units of m.

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

Answer 2

Part One of Three
Dx=20 meters
Dy=3 meters (Don't need to know that for this part of the question)
g=-9.8 m/s^2
Vix=17 m/s
Vi in this formula is cos(35)*17
Formula: Dx=Vit
20=cos(35)*17
20=13.93t
20/13.93= 1.44
T=1.44 seconds!

Part Two of Three
Viy=sin(35)*17=9.75 m/s
T (found in Part 1 of 3)=1.44 seconds
Dy=3 meters (not needed until end of problem)
a=g or (-9.8 m/s^2)
Formula Used: Dy=Viyt+1/2at^2
Ignore Dy value of 3 for now
Dy=9.75*1.44+1/2(-9.8)1.44^2
Dy=14.04-10.16=3.88
Now dy=3 meters
3.88-3=0.88 meters above the fence.

Part Three of Three
Formula Used= Vfy=Viy+ay*t
Viy=9.75
A=-9.8m/s^2 or g
T=1.44 seconds
Vfy=9.75+(-9.8)*1.44
-(negative)*+(positive)=-(negative)
Vfy=9.75-14.11
Vfy=-4.36