Question

A ball thrown from the edge of a building hits the ground at an angle of 60°with the horizontal, 25 m from the building and 2.5s after it is thrown.

(take g = 10m/s² and $\sqrt{3}$ = 17)

The direction of velocity with the horizontal with which the ball was thrown, is

• A. θ = tan^-14/5
• B. θ = tan^-13/5
• C. θ = tan^-12/5
• D. θ = tan^-17/10

Answer 1

Answer: (A)

θ = tan^-14/5

Answer 2

For horizontal motion, as component of velocity in this direction is not changed with time

t = 2.5 s = 25m/u_x ⇒ u_x =10 m/s

Also at the moment of landing, velocity is V, then

V cos 60^o = u_x ⇒ V = $\frac{10}{(1/2)}$ = 20 m/s

and V_y = V sin 60⁰ = $20\frac{\sqrt{3}}{2}$= $10\sqrt{3}$ m/s

For the vertical motion, using v = u + at and taking upward direction as positive

- $10\sqrt{3}$ = u_y + (-10)(2.5)

⇒ u_y = 25 - $10\sqrt{3}$ = 8 m/s, upward.

Angle of projection tan θ = $\frac{u_{y}}{u_{x}} = \frac{8}{10} = \frac{4}{5}$