Question

A tennis ball is dropped from a height of 4m and it rebounds to aheight of 3 m . if the ball was in contact with floor for o.o1s . find average acceleration during contact.

Answer 1

Final Answer :
$1669.02 \: m \div {s}^{2}$

Steps:
1) Convention :
Upward direction + ,
Downward Direction - :

We observe that,
Initial velocity when ball strikes the ground can be given as
$u= - \sqrt{2g \times 4 } = - \sqrt{80} \\ = - 8.94 \: m \div s$
Final Velocity,
$v = \sqrt{2g \times 3} = \sqrt{60} \\ = 7.75 \: m \div s$

3) Time of contact at ground,
$\delta \: t \: = 0.01s$
By Impulse Momentum Theorem,

Assumimg Impulsive Force to be constant Throughout t = 0.01s
$f \: \delta t = \delta p$

Ft = m(v-u)
=> F/m = (v-u) /t
=> a =[ 7.75-(-8.94))]/0.01 = 1669 m/s^2

Hence, Average Acceleration is 1669 m/s^2

Integrating Both sides,

Answer 2

Answer: u = sqrt(2gH)

= -sqrt(2 × 9.8 × 4)

v = -sqrt(2gh)

= sqrt(2 × 9.8 × 3)

Average acceleration = (v - u) / t

= (v - (-u)) / t

= (v + u) / t

= [sqrt(2 × 9.8 × 3) + sqrt(2 × 9.8 × 4)] / 0.01

= 1652.25 m/s^2

Average acceleration during contact is 1652.25 m/s²