Question

A ball is thrown up with velocity of 10m/sec, it reached to a certain height and started returning, Find the velocity of the ball after covering 4m in return journey. (Mass of the ball=100gm, acceleration due to gravity= 10m/ 2 ).​

Answer 1

Answer:

From third equation of motion ,v

2

=u

2

−2gh

0=u

2

−2g(h/2)

h=

g

u

2

=

10

10

2

=10m

Thus, ball rises to 10m

Answer 2

Answer:

The velocity of the ball after covering 4 m in return journey is 8.94 m/s.

Explanation:

Given information:

• Initial upward velocity of the ball $u_1=10\text{ m/s}$.

At maximum height the velocity of the ball will be zero i.e.$v_1=u_2=0$. Here, $u_2$ is the initial downward velocity.

The velocity of the ball after covering 4 m in return journey is given as,

$v_2=\sqrt{u_2^2+2gh}$

Substituting all known values,

[tex]v_2=\sqrt{(0\text{ m/s})^2+2\times(10 \text{ m/s})\times(4\text{ m})}\\ \approx8.94\text { m/s}[/tex]