A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ . [Given ] [2023]

Question

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ ${ms}^{- 2}$ . [Given $g = 10$ ${ms}^{- 2}$ ] [2023]

Smart Achievers · Accepted Answer

(120)

$v_{i} = \sqrt{2 g h_{i}} = \sqrt{2 \times 10 \times 9.8} ↓ = 14 m/s ↓$

$v_{f} = \sqrt{2 g h_{f}} = \sqrt{2 \times 10 \times 5} ↑ = 10 m/s ↑$

$| {\vec{a}}_{avg} |$ $=$ $| \frac{Δ \vec{v}}{Δ t} |$ $= \frac{24}{0.2} = 120 {m/s}^{2}$

Solution

Q.
A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ ${ms}^{- 2}$ . [Given $g = 10$ ${ms}^{- 2}$ ] [2023]

Ans.
(120)

$v_{i} = \sqrt{2 g h_{i}} = \sqrt{2 \times 10 \times 9.8} ↓ = 14 m/s ↓$

$v_{f} = \sqrt{2 g h_{f}} = \sqrt{2 \times 10 \times 5} ↑ = 10 m/s ↑$

$| {\vec{a}}_{avg} |$ $=$ $| \frac{Δ \vec{v}}{Δ t} |$ $= \frac{24}{0.2} = 120 {m/s}^{2}$

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Solution

Q. A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ ms-2. [Given g=10 ms-2] [2023]

Ans. (120) vi=2ghi=2×10×9.8 ↓=14 m/s↓ vf=2ghf=2×10×5 ↑=10 m/s↑ |a→avg|=|Δv→Δt|=240.2=120 m/s2

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Q.
A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height of 5.0 m. Ball comes in contact with the floor for 0.2 s. The average acceleration during contact is ________ ${ms}^{- 2}$ . [Given $g = 10$ ${ms}^{- 2}$ ] [2023]

Ans.
(120)

$v_{i} = \sqrt{2 g h_{i}} = \sqrt{2 \times 10 \times 9.8} ↓ = 14 m/s ↓$

$v_{f} = \sqrt{2 g h_{f}} = \sqrt{2 \times 10 \times 5} ↑ = 10 m/s ↑$

$| {\vec{a}}_{avg} |$ $=$ $| \frac{Δ \vec{v}}{Δ t} |$ $= \frac{24}{0.2} = 120 {m/s}^{2}$