Question

A boy throws a stone upward with the velocity of 60m/s. (a) how long will it take to reach the maximum height? (b)what is the maximum height reached by the ball? (c)how long will it take to reach the ground?

Answer 1

Given:

Initial velocity of stone, u=60 m/s

To Find:

(a) Time taken by stone to reach the maximum height

(b) Maximum height reached by the stone

(c) Time taken by stone to reach the ground

Solution:

We know that,

• When a body is thrown vertically upwards, then velocity of body at highest point is 0

• According to first equation of motion for constant acceleration,

$\purple{\boxed{\bf{v=u+at}}}$

• According to second equation of motion for constant acceleration,

$\orange{\boxed{\bf{s=ut+\dfrac{1}{2}at^{2}}}}$

• According to third equation of motion for constant acceleration,

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

where,

v is final velocity

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.

$\rule{190}{1}$

We have to consider two cases:

Case-1: When object is going upward

Let the time taken by stone to reach the highest point be t, final velocity be v and s be the displacement of stone

On applying first equation of motion for upward motion of stone, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=60+(-g)t}$

$\longrightarrow\rm{-60=-gt}$

$\longrightarrow\rm{-60=-10t}$

$\longrightarrow\rm{-10t=-60}$

$\longrightarrow\rm{t=\dfrac{-60}{-10}}$

$\longrightarrow\rm\green{t=6\ s}$

Again, on applying third equation of motion for upward motion of object, we get

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

$\longrightarrow\rm{(0)^{2}-(60)^{2}=2(-g)s}$

$\longrightarrow\rm{-3600=2(-10)s}$

$\longrightarrow\rm{-3600=-20s}$

$\longrightarrow\rm{-20s=-3600}$

$\longrightarrow\rm{s=\dfrac{-3600}{-20}}$

$\longrightarrow\rm\green{s=180\ m}$

$\rule{190}{1}$

Case-2: When stone is going downward

Let the time taken by stone to reach the ground be t', initial velocity be u' and s be the displacement of the stone

On applying second equation of motion for downward motion of stone, we get

$\longrightarrow\rm{s=u't'+\dfrac{1}{2}a(t')^{2}}$

$\longrightarrow\rm{-180=0(t)+\dfrac{1}{2}(-g)(t')^{2}}$

$\longrightarrow\rm{-180=\dfrac{1}{\cancel{2}}(\cancel{-10})(t')^{2}}$

$\longrightarrow\rm{-180=-5(t')^{2}}$

$\longrightarrow\rm{-5(t')^{2}=-180}$

$\longrightarrow\rm{(t')^{2}=\dfrac{-180}{-5}}$

$\longrightarrow\rm{(t')^{2}=36}$

$\longrightarrow\rm{t'=\sqrt{36}}$

$\longrightarrow\rm{t'=6\ s}$

Let the total time taken by stone to reach the ground be T

So,

$\longrightarrow\rm{T=t+t'}$

$\longrightarrow\rm{T=6+6}$

$\longrightarrow\rm\green{T=12\ s}$

$\rule{190}{1}$

Hence,

Time taken by stone to reach the maximum height is 6 s.

Maximum height reached by the stone is 180 m.

Time taken by stone to reach the ground is 12 s.