Question

a ball is dropped on the floor from a height of 10m. it rebounds to a height of 2.5m. if the ball is in contact with ground for 0.02secs its average velocity is

1) 2100 ms^-2
2) 1050ms^-2
3) 4200ms^-2
4) 9.8ms^-2

Answer 1

$\Large{\underline{\underline{\bold{☆Given:}}}}$

• Height the ball dropped from is 10 metres.
• Height Rebounded = 2.5m
• Time the ball is in contact with surface of ground = 0.02 seconds.

$\Large{\underline{\underline{\red{\bold{➸To\;Find:}}}}}$

• The Average Velocity of the ball.

$\Large{\underline{\underline{\bold{\color{salmon}๛Required\; Response:}}}}$

Let, The Velocity with which the ball hits the ground be 'u'. Then,

• $\pink{\bold{u² = 2gh}}$

${u}^{2} = 2 \times 9.8 \times 10 \\ {u}^{2} = 196 \\ u = \sqrt{196} \\ u = \sqrt{ ({14})^{2} } \\ u = 14$

.:. $\bold{u = 14m/sec}$

If v be the velocity with which the ball rebounds. Then,

• $\pink{\bold{v² = 2gh_2}}$

Where, $\sf{h_2}$ is the rebounds height.

${v}^{2} = 2 \times 9.8 \times 2.5 \\ {v}^{2} = 49 \\ v = \sqrt{49} \\ v = \sqrt{ ({7})^{2} } \\ v = 7$

.:. $\bold{v = 7m/sec}$

• $\pink{\bold{∆v = (v-u)}}$

$\bold{(7m/sec)-(-14m/sec)}$

$\bold{21m/sec}$

• $\pink{\bold{a = \frac{∆v}{∆t}}}$

$\frac{21}{0.01} \\ \frac{21}{ \frac{1}{100} } \\ 21 \times 100 \\ 2100$

The Average Velocity of the ball

• ☞ $\Large\boxed{\blue{2100\;ms^{-2}}}$

$\boxed{\tt{@MrSovereign}}$

Hope It Helps You ✌️