Question

A ball is gently dropped from a height of 20m on riding of its velocity increases uniformly @ 10m per second square with what velocity will it strike the ground

Answer 1

ball is thrown from the height so initial velocity is given and final velocity will zero by Newton's second law of motion

Answer 2

AnswEr:

Given,

• Initial velocity, u = 0 ( as it is dropped )
• Height, h = 20 m
• Acceleration , a = 10 m $\sf{s}^{-2}$
• Final Velocity, v = ?
• Time, t = ?

$\bold{\large{\boxed{\sf{\pink{v^2-u^2+2as}}}}}$

Here, s = h

$\tt \: {v}^{2} = {u}^{2} + 2as \\ \\ \tt = 0 + 2 \times 10 \times 20 \\ \\ \tt = 0 + 400 = 400 \\ \\ \tt \leadsto \: v = \sqrt{400} = v = 20m \: s { }^{ - 1}$

• Time = ?

= Using first equation of motion

$\tt v= u + at \\ \\ \tt = \frac{v - u}{a} = t \\ \\ \tt = \frac{20 - 0}{10} = \frac{2 \cancel0}{1 \cancel0} \\ \\ \tt = 2s$

Therefore, time will be 2 s.