Question

A boy on the edge of a vertical cliff 20-m high throws a stone horizontally outwards with a speed of 20 m/s. It strikes the ground at what horizontal distance from the foot of the cliff? (Use g = 10 m/s2.)

Answer 1

Answer:

Horizontal distance from the foot of the cliff is 40 m

Explanation:

Given :

Speed of stone $v = 20\frac{m}{s}$

Vertical distance $d = 20$ m

Gravitational acceleration $g = 10$ $\frac{m}{s^{2} }$

From the kinematics equation,

  $d = v_{o} t + \frac{1}{2} at^{2}$

Where $v_{o} =$ initial velocity ( $v_{o} = 0$ )

So time take to fall down,

  $t = \sqrt{\frac{2d}{a} }$

  $t = \sqrt{4}$

  $t = 2$ sec

Now find horizontal distance travel by stone,

 $x = vt$

 $x = 20 \times 2$

 $x = 40$ m

Therefore, the horizontal distance travel by stone is 40 m