Question

ball is thrown vertically upward. It has speed of 10m/s when it is reached one fourth of its maximum height. How high does the ball rise? (Taking g=10m/s 2

Answer 1

Given:

Velocity of ball at one-fourth of maximum height attained by ball= 10 m/s

To Find:

Maximum height attained by ball

Solution:

We know that,

• When a body is thrown vertically upwards, then velocity of body at highest point is 0
• According to third equation of motion for constant acceleration,

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

$\rule{190}{1}$

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.

$\rule{190}{1}$

Let the maximum height attained by the ball be H, initial velocity of ball be u

So, on applying third equation of motion on ball for the interval in which ball attained one-fourth of the maximum height, we get

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

$\longrightarrow\rm{(10)^{2}-u^{2}=2(-g)\bigg(\dfrac{H}{4}\bigg)}$

$\longrightarrow\rm{100-u^{2}=2(-10)\bigg(\dfrac{H}{4}\bigg)}$

$\longrightarrow\rm{100-u^{2}=\dfrac{-20H}{4}}$

$\longrightarrow\rm{100-u^{2}=-5H}$

$\longrightarrow\rm{100+5H=u^{2}}$

$\longrightarrow\rm{u^{2}=100+5H}------(1)$

$\rule{190}{1}$

Now, on applying third equation of motion on ball for its journey in which it attains maximum height, we get

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

$\longrightarrow\rm{(0)^{2}-u^{2}=2(-g)H}$

From (1), we get

$\longrightarrow\rm{(0)^{2}-(100+5H)=2(-10)H}$

$\longrightarrow\rm{-(100+5H)=-20H}$

$\longrightarrow\rm{20H=100+5H}$

$\longrightarrow\rm{20H-5H=100}$

$\longrightarrow\rm{15H=100}$

$\longrightarrow\rm{H=\dfrac{100}{15}}$

$\longrightarrow\rm\green{H=6.67\ m}$

Hence, the maximum height attained by ball is 6.67 m.