Question

A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground.

Answer 1

Given:
Speed of the ball, u_x = 20 m/s
Height from which the ball is dropped, h = 100 m

(a) Let t be the time taken by the ball to reach the ground.
Using the equation of motion, we have:
$h=u_{y}t+\frac{1}{2}g{t}^2$
Here,
Acceleration of gravity, g = 9.8 ms⁻²
Vertical component of velocity, u_y = 0
$$\begin{gathered} \therefore t=\sqrt{\left(\frac{2h}{g}\right)} \\ \Rightarrow t=\sqrt{\frac{2\times 100}{9.8}}=4.5\mathrm{s} \end{gathered}$$
Therefore, the time required by the ball to reach the ground is 4.5 seconds.

(b) Horizontal distance travelled by the ball:
x = u_xt = 20 × 4.5 = 90 m

(c) We know that horizontal velocity remains constant throughout the motion of the ball.
At A, v_x = 20 m/s.
v_y = u + gt = 0 + 9.8 × 4.5
⇒v_y = 44.1 m/s
Resultant velocity:
$$\begin{gathered} v_{\mathrm{r}}=\sqrt{{\left(44.1\right)}^2+{\left(20\right)}^2}\approx 49\mathrm{m}/\mathrm{s} \\ \tan \beta =\frac{v_y}{v_x}=\frac{44.1}{20}=2.205 \\ \Rightarrow \beta ={\tan }^{-1}\left(2.205\right)=66° \end{gathered}$$
Therefore, the ball strikes the ground with a magnitude of velocity 49 m/s and the direction at an angle of 66° with the ground.