Question

A hiker stands on the edge of a cliff 490m above the ground and throws a stone horizontally with an initial speed of 15m/s. Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground (Take ${\text{g = 9}}.{\text{8m/}}{{\text{s}}^{\text{2}}}$ ).

Answer 1

The ball is dropped from height so it is projection from a height. Motion of a stone may be considered as the superposition of the two independent motions. Taking the equation of motion in vertical and horizontal direction will help in solving further.

Formula used We will start by solving equations in vertical and horizontal direction.
We will also be using the equations of motion.
${\text{s = ut + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{t}}^{\text{2}}}$
${{\text{v}}^{\text{2}}}{\text{ = }}{{\text{u}}_{}}^{\text{2}}{\text{ + 2gs}}$

Complete Answer:
Here, we already know that there will be no change in magnitude of the horizontal component. Only the vertical component will change.
So, we start in the horizontal direction:
${\text{u = 15m/s}}$
Vertical motion with constant acceleration:
${\text{a = g = 9}}.{\text{8m/}}{{\text{s}}^2}$
The height of the cliff is considered to be ${\text{s}}$ .
Also, let us consider ${{\text{u}}_{\text{y}}}$ be the vertical component of the velocity of the projection.
Since, the stone is thrown horizontally, ${{\text{u}}_{\text{y}}}$ =0
The stone hits the ground after t seconds of projection,
So, ${\text{s = }}{{\text{u}}_{\text{y}}}{\text{t + }}\dfrac{{\text{1}}}{{\text{2}}}{\text{g}}{{\text{t}}^{\text{2}}}$
Time required by the stone to reach ground:
${\text{t = }}\sqrt {\dfrac{{{\text{2 \times 490m}}}}{{{\text{9}}{\text{.8m/}}{{\text{s}}^{\text{2}}}}}} {\text{ = 10sec}}$
${\text{t = }}\sqrt {\dfrac{{{\text{2}} \times {\text{490m}}}}{{{\text{9}}.{\text{8m/}}{{\text{s}}^{\text{2}}}}}} {\text{ = 10sec}}$
Similarly, we will find the vertical direction:
${{\text{v}}_{\text{y}}}^{\text{2}}{\text{ = }}{{\text{u}}_{\text{y}}}^{\text{2}}{\text{ + 2gs}}$
$\Rightarrow {{\text{v}}_{\text{y}}}^{\text{2}} = {0^2} + 2 \times 9.8 \times 490$
On solving, we get
$\Rightarrow {{\text{v}}_{\text{y}}}^{\text{2}}{\text{ = 9604}}$
$\Rightarrow {{\text{v}}_{\text{y}}}{\text{ = 98m/s}}$
Since, the horizontal component remains constant.
${{\text{v}}_{\text{x}}}{\text{ = }}{{\text{u}}_{\text{x}}}{\text{ = 15m/s}}$
So, the final speed of the stone,
${\text{v = }}\sqrt {{{\text{v}}_{\text{x}}}^{\text{2}}{\text{ + }}{{\text{v}}_{\text{y}}}^{\text{2}}}$
$\Rightarrow {\text{v = }}\sqrt {{{15}^2} + {{98}^2}}$
On solving the above equation,
$\Rightarrow \sqrt {9829}$
$\Rightarrow {\text{99}}.{\text{14m/s}}$
Thus, the final speed with which the stone hits the ground is 99.14 m/s.

Note:
The common mistake during the evaluation is doing the evaluation. It should be done carefully, as cases are different in case of ground-to-ground projection and projection from a height.
Also, acceleration due to gravity $\left( {\text{g}} \right)$ always acts vertically downwards and there is no acceleration in horizontal direction unless mentioned.
Take the value of $\left( {\text{g}} \right)$ to be $9.8m/{s^2}$ if not mentioned in the question. Generally, the value is considered to be $10m/{s^2}$ for the sake of calculation.