Question

8. A tennis ball is dropped on the floor from a height of
10 m. It rebounds to a height of 5 m. The ball was in
contact with the floor for 0.01 s. What was its average
acceleration during the contact ? (g=10m/s2)
(1) 800 m/s2
(2) 200 m/s2
(3) 1600 m/s2
(4) 2400 m/s2​

Answer 1

Given :

A tennis ball is dropped on the floor from a height of 10m. It rebounds to a height of 5m.

Time of contact = 0.01 s

To Find :

Average acceleration of ball during the contact.

Solution :

❒ We know that, average acceleration is measured as the rate of change of velocity.

Third equation of kinematics;

• v² - u² = 2gH

» v denotes final velocity

» u denotes initial velocity

» g denotes acceleration

» H denotes height

A] Velocity of ball just before impact :

➠ v₁² - u₁² = 2gH₁

• Initial velocity = zero

➠ v₁² = 2gH₁

➠ v₁ = √2gH₁

B] Velocity of ball just after impact :

➠ v₂² - u₂² = 2gH₂

• Final velocity = zero

➠ v₂² = 2gH₂

➠ v₂ = √2gH₂

★ Acceleration of ball :

$\sf:\implies\:a=\dfrac{v_2-(-v_1)}{t}$

• Negative sign shows opposite direction.

$\sf:\implies\:a=\dfrac{\sqrt{2gH_2}+\sqrt{2gH_1}}{t}$

$\sf:\implies\:a=\dfrac{\sqrt{2(10)(10)}+\sqrt{2(10)(5)}}{0.01}$

$\sf:\implies\:a=\dfrac{\sqrt{200}+\sqrt{100}}{t}$

$\sf:\implies\:a=\dfrac{14.1+10}{0.01}$

$:\implies\:\underline{\boxed{\bf{\purple{a\approx2400\:ms^{-2}}}}}$

Answer 2

Question:

8. A tennis ball is dropped on the floor from a height of

10 m. It rebounds to a height of 5 m. The ball was in

contact with the floor for 0.01 s. What was its average

acceleration during the contact ? (g=10m/s2)

(1) 800 m/s82

(2) 200 m/s2

(3) 1600 m/s2

(4) 2400 m/s2

Answer:

4) 2400 m/s^2