Question

A stone is thrown horizontally from a cliff 30m high with an initial speed of 20m/s . how far from the cliff does the stone strike the ground

Answer 1

Given :-

Initial Velocity of stone = u = 20 ms-¹

Height of cliff = h = 30 m

As considering the vertical initial velocity of body as zero.

S = ut + 1/2 gt²

30 = 0 + 1/2 × 10t²

t = √(2×30)/10

t = 2.4 s

Again,

As given the horizontal velocity of stone is constant hence,

d = vt

d = 20 × 2.4

d = 48 m

Hence,

The distance it will strike the ground = d = 48 m

Answer 2

Answer :

Height of cliff = 30m

Initial speed = 20m/s

We have to find range of stone$.$

★ First of all we need to find time of flight.

In such kind of height to ground projectile motion, Horizontal component of initial velocity determines range of projectile and vertical component of initial velocity determines time of flight.

• Horizontal component of velocity remains constant throughout the motion as no acceleration acts along horizontal direction.
• Vertical component of velocity changes with time as acceleration due to acts in downward direction.

➝ t = √2h/g

➝ t = √2(30)/10

➝ t = √60/10

➝ t = √6

➝ t = 2.45 s

★ Calculation of horizontal range :

➝ R = u × t

• u denotes initial velocity
• t denotes time of flight
• R denotes range

➝ R = 20 × 2.45

➝ R = 49 m