Motion — Question 7

Given,

Initial velocity (u) = 0

Distance travelled (s) = 20 m

Acceleration (a) = 10 ms^-2

As per the third law of motion,

v² – u² = 2as

Substituting we get,

v² - 0 = 2 x 10 x 20

⇒ v² = 400

⇒ v = $\sqrt{400}$

⇒ v = 20 ms^-1

Hence, ball strikes the ground with a velocity of 20 ms^-1

As per the first equation of motion,

v = u + at

or

t = $\dfrac{\text{v - u}}{\text{a}}$

Substituting we get,

t = $\dfrac{20-0}{10}$ = 2 seconds

∴ the ball reaches the ground after 2 seconds.