KnowledgeBoat Logo

Science

A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of 10 ms-2, with what velocity will it strike the ground? After what time will it strike the ground?

Motion in One Dimension

1 Like

Answer

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 = 400\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 = v - ua\dfrac{\text{v - u}}{\text{a}}

Substituting we get,

t = 20−010\dfrac{20-0}{10} = 2 seconds

∴ the ball reaches the ground after 2 seconds.

Answered By

1 Like

Related Questions