Question

A balloon is moving with a velocity of 5m/s in upward direction. At t=0 balloon is at ground .At t=10s ball is left gently from balloon .Find speed of ball at ground
a)32m/s
b)16m/s
c)49m/s
d)20m/s

Answer 1

Given:

➺ At t= 0

Ball was at ground

Constant velocity of balloon= 5 m/s

➺ At t= 10 s

Ball is left gently from balloon

To Find:

Speed of ball at ground

Solution:

We know that,

• $\pink{\boxed{\bf{Distance=Speed\times Time}}}$
• According to third equation of motion for constant acceleration,

$\purple{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

u is initial velocity

a is acceleration

s is displacement

t is time taken

$\rule{190}{1}$

Reference taken here:

All displacements, velocities, forces and accelerations acting in upward direction are taken positive.

All displacements, velocities, forces and accelerations acting in downward direction are taken negative.

$\rule{190}{1}$

For the upward motion of balloon, let the displacement of balloon at t= 10s be d

So,

$\longrightarrow\rm{d=Speed\times time}$

$\longrightarrow\rm{d=5\times10}$

$\longrightarrow\rm{d=50\:m}$

So, balloon attained the height of 50 m after 10s

$\rule{190}{1}$

Now, ball is dropped from balloon at t= 10s

So, at t= 10s

Height of  ball, d= 50 m

Initial velocity of ball, u= 5 m/s

Let the velocity of ball at ground be 'v'

On, applying third equation of motion for the downward motion of ball, we get

$\longrightarrow\rm{v^{2}-u^{2}=2as}$

$\longrightarrow\rm{v^{2}-(5)^{2}=2(-g)d}$

$\longrightarrow\rm{v^{2}-25=2(-10)(-50)}$

$\longrightarrow\rm{v^{2}-25=2(500)}$

$\longrightarrow\rm{v^{2}-25=1000}$

$\longrightarrow\rm{v^{2}=1000+25}$

$\longrightarrow\rm{v^{2}=1025}$

$\longrightarrow\rm{v^{2}=1025}$

$\longrightarrow\rm{v=\sqrt{1025}}$

$\longrightarrow\rm\green{v=32\:m/s\:(approx.)}$

Hence, the speed or velocity of ball at the ground is 32 m/s and the correct option is (a).