Question

a ball is thrown from the top of one building towards a nearby very tall building 50m away. the initial velocity of the ball is 20 m/s at 40 degrees above thehorizontal. how far above or below its original level will the ball strike the opposite wall?

Answer 1

The distance between two buildings x = 50m

the initial velocity v0 = 20m/s

then the horizontal comoponent vx = v0 cosθ

= (20)(cos40)

= 15.32 m/s

the time interval to reach the building

x = vx t

then t = x/vx = 50/15.32

= 3.27 sec

the maximum height reached h = (vsinθ)^2 /2g

= (20sin40)^2 /2(9.8)

= 8.4m

the vertical displacement

y= vy t- 1/2 gt^2

= (20 sin40)(3.27)-(0.5)(9.8)(3.26)^2

= 41.9- 52 = -10.27 m

Therefore the ball will be at10.27 mbelow from its orginal level