91Ó°ÊÓ

The velocity contains two components, one is horizontal component and another one is vertical.

According to the newton’s laws of motion,

$s=ut+\frac{1}{2}a{t}^2$

Where, s, u, t, a, are displacement, initial velocity, time, and acceleration.

For the vertical and horizontal motion, the velocity component will berole="math" localid="1663739048893" $\mathrm{u}\mathrm{²õ¾\pm ²Ôθ}\mathrm{and}\cos \mathrm{θ}$ respectively.

Weight of ball =$2.7\mathrm{Kg}$

Initial speed = 20m/s

Cliff height = 45m

Woman’s speed during running away from the base of cliff = 6.0m/s

For the horizontal speed,

$$\begin{gathered} u_h=u\cos \mathrm{θ} \\ \cos \mathrm{θ}=\frac{6.00m/s}{20.00m/s} \\ \mathrm{θ}=72.5° \end{gathered}$$

The ball is in air for 5.55s so the horizontal distance run by her,

$$\begin{gathered} \mathrm{s}=\left(6.00\mathrm{m}/\mathrm{s}\right)\left(5.55\mathrm{s}\right) \\ \mathrm{s}=33.3\mathrm{m} \end{gathered}$$

When the ball trajectory will be viewed by a person at the rest on the ground the ball moves in a parabolic curve.

When the ball trajectory will be viewed by a the runner the motion of ball will be only vertical.