Question

a small ball drop from a height 100m on a floor .the ball rebounce to height 20 m calculated average acceleration during the constant if the ball was in constant of floor for 0.02sec takes g=10m/S2 ​

Answer 1

Answer:

When, h=20m,

==>> V^2=u^2 - 2gh or

==>> v =√2gh (downwards ,so negative)

→ When ball rebounds, h=10m

→ So ,0^2 =v^2–2gh

» Or v= √2gh (upwards,so positive)

≈>Now,acceleration of ball,

a =v(upwards)-v(downward)÷t

=√2×9.8×10-(-√2×9.8×20)÷0.1

=338 ms^-1