A stone is dropped from a height of 20 m.
(i) How long will it take to reach the ground ?
(ii) What will be its speed when it hits the ground ? $(g = 10 m / s^{2})$

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Solution

Given :

Height h = 20

Initial velocity u = 0m/s

Acceleration due to gravity, $(g = 10 m / s^{2})$

(i) Time to reach the ground from 2nd equation of motion

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

$h = u t + \frac{1}{2} g t^{2}$

$20 = 0 \times t + \frac{1}{2} \times 10 \times t^{2}$

$20 = 5 \times t^{2}$

$t^{2} = 20 / 5 = 4$

t = √4= 2 sec

(ii) Speed when it hits the ground t = 2sec from 1st equation of motion

v = u + gt

= 0 + 10 × 2 = 20 m/s

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Similar questions

g = 10 m / s^{2}

1.A stone falls from a building and reaches the ground in 2.5 seconds latter . How high is the building?

2. A stone dropped from a height of 20meter

i) How long it will take to reach the ground.

ii)what will be it's speed when it hits the ground.

10 m / s

40 m

(t a k e g = 10 m / s^{2})

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