Question

A ball is dropped from the top of a building $100m$ high. At the same instant another ball is thrown upwards with a velocity of $40m{s}^{-1}$ from the building. The two balls will meet after.

• A. $5s$
• B. $2.5s$
• C. $2s$
• D. $3s$

Answer 1

Answer: (B)

$2.5s$

Given,

$H=100m$

For ball 1: $u=0;g=10m/s$

So, $h_1=u_1+\frac{1}{2}g{t}^2=0\times t+\frac{1}{2}(10{)}^2{t}^2=5t^2$ ----------(1)

For ball2: $u=40m/s;a=-10m/s$

So, $h_2=u_2+\frac{1}{2}a{t}^2=40\times t+\frac{1}{2}(-10{)}^2{t}^2=40t-5t^2$ ---------(2)

$h=h_1+h_2$

$h=5t^2+40t-5t^{2}h=100m$

$100=40t$

$t=\frac{5}{2}=2.5s$