Question

A stone us dropped from height of 80m. Find- Time of fall, velocity after 2 seconds of free fall and velocity with which it hits the ground.​

Answer 1

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given :-

h or s = 80m

u = 0m/s ( at rest )

g = 10m/s²(gravitational acceleration)

by second equation of motion for gravitation

$s = ut + \frac{1}{2} g{t}^{2} \\ \implies80 = 0 + \frac{1}{2} \times 10 {t}^{2} \\ \implies \: {t}^{2} = \frac{80 \times 2}{10} \\ \implies {t}^{2} = \frac{160}{10} \\ \implies \: t = \sqrt{16} \\ \implies \: 4sec$

so after 4 second the ball hits the ground..

now for finding velocity of ball after 2 seconds..

we will use first equation of motion for gravitation..

$v = gt + u \: \\ \implies \: v = 10 \times 2 + 0 \\ \implies \: v = 20m {s}^{ - 1}$

so velocity of ball after 2 seconds is 20m/s

now for finding the velocity of ball with which it hits the ground

we will use first equation of motion for gravitation

$v = gt + u \\ \implies \: v = 10 \times 4 + 0 \\ \implies v = 40m {s}^{ - 1}$

so with velocity of 40m/s ball hits the ground..

hence,

• total time taken by the ball is 4 seconds
• velocity of ball after 2 seconds is 20m/s
• velocity of ball with which its hits the ground is 40m/s

Answer 2

ABOVE ANSWER IS CORRECT