Question

A ball is dropped on the floor from a height of 10 m . It rebounds to a height of 2.5m. If the ball is in the contact with the floor for 0.01 sec., the average accleration during contact is ________?​

Answer 1

See the attachment........

Answer 2

Given :-

Height dropped = H = 10 m

Height rebounds = h = 2.5 m

Time of contact = t = 0.01 s

Concept :-

When ball is dropped then it has final velocity only it doesn't have initial velocity but when rebounds back it has only initial velocity and it doesn't have final velocity.

So Final velocity while dropped.

v = √2gH

v = √2×9.8×10

v = √196

v = 14 ms-¹

Again,

Initial Velocity when ball rebounds.

u = √2gh

u = √2×9.8×2.5

u = √49

u = 7 ms-¹

${a}_{avg}$ = ∆v/∆t

${a}_{avg}$= (14 - 7)/0.01

${a}_{avg}$ = 700 ms-²