Question

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

Answer 1

v^2=u^2+2as

u=0 a=10 s=20

So v(final velocity)=√(2as)=√(2*10*20)=20m/s

And u+at=v

t=v/a=20/10=2s

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Answer 2

Velocity is a quantity with both magnitude and direction, and a change in velocity can either or both of these factors.

From the given

Ball is being dropped, therefore, initial velocity “a” = 20 m

u = 0 m/s

a = 10 m/s2

According to the third law of thermodynamics,

V^2 = u^2 + 2as

Substitute all the given values, then,

V^2 = 0^2+2×10×20

= 400

V = 20

To find the time “t”,

v = u+at

20 = 0+10t

t = 20/10

t = 2sec