Question

A ball is thrown vertically upward with a velocity of 5 m/s. it uniformly accelerates at rate of 10 m/s ,stops after reaching a certain height and begins to fall back downward . calculate the height attained by ball before it falls back

Answer 1

Given:

A ball is thrown vertically upward with a velocity of 5 m/s. it uniformly accelerates at rate of 10 m/s ,stops after reaching a certain height and begins to fall back downward .

To find:

Height attained by ball before it falls back.

Calculation:

Acceleration of the ball is opposite to the direction of velocity. This causes the ball to stop after reaching a certain height.

Considering the Acceleration to be constant, we will apply the equation of Kinematics to solve this problem :

$\therefore \: {v}^{2} = {u}^{2} + 2as$

$= > {(0)}^{2} = {(5)}^{2} + 2 \times ( - 10) \times h$

$= > 20h = 25$

$= > h = \dfrac{25}{20}$

$= > h = 1.25 \: m$

So , the maximum height reached is 1.25m.