Will the Ball Hit the Fence or Pass Above It?

[logglypop]

A ball of mass 0.5 kg, initially at rest, is kicked directly towards a fence from a point 32 meters away. The velocity of the ball as it leaves the kicker's foot is 20 meters per second at an angle of 37 degrees above the horizontal. The top of the fence is 2.5 meteres high. The kicker's foot is in contact with the ball for 0.05 second. The ball hits nothing while in flight and air resistance is negligible.

Will the ball hit the fence? If so, how far below the top of the fence will it hit? If not, how far above the top of the fence will it pass?



time = distance/velocity
time = 32 m / horizontal velocity
time = 32 m / ((cos 37) * 20) m/s
time = 32 m / ((4/5) * 20) m/s
time = 2 seconds

d = vt + .5at^2
d= 20*2 + .5 (-9.8) 2^2
d=20.4

20.4 - 2.5 = 17.9
the ball won't hit the fence. 17.9m above the fence

need confirm please

 

[hage567]

d= 20*2 + .5 (-9.8) 2^2

You didn't take the angle into account here.

 

[logglypop]

the i guess the
V=20cos37
V=16
d= 16*2 + .5 (-9.8) 2^2
d=13.5m
so the ball 13.5 above the fence?

 

[hage567]

No, 20cos37 is the x component of the initial velocity. You need to find the y component since you are trying to find out the height of the ball when it reaches the fence.

 

[logglypop]

No, 20cos37 is the x component of the initial velocity. You need to find the y component since you are trying to find out the height of the ball when it reaches the fence.

thanksV=20sin37
V=12
d= 12*2 + .5 (-9.8) 2^2
d=4.4m

 

[hage567]

Yes, that's the height of the ball above the ground when it reaches the fence.