Question

As shown in the figure, a ball is thrown from the top of one building toward a tall building 50 m away. The initial velocity of the ball is 20 m/s at 40 degree above the horizon. How far above or below its original level will the ball strike the opposite wall. What is the velocity of the ball at the impact point (landing point)?

Answer 1

The x and y component of velocity are:

vx = v cos $\theta$ = 20 x cos40 = 15.32 m/s

vy = v sin $\small \theta$ = 20 x sin40 = 12.86 m/s

We know that, horizontal distance is:

D = v x * t => t = D/vx = 50/15.32 = 3.26 s

from eqn of motion

S = ut + 1/2 at^2

y = 12.86 x 3.26 - 0.5 x 9.8 x 3.26^2 = -10.15 m

Hence, y = -10.15 m

b) v = u + at

v = 12.86 + 9.8 x 3.26 = 44.81 m/s

Hence, v = 44.81 m/s