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QJEEM23Phf0fb: JEE Main 2023 Physics - Class 11 JEEM23Phf0fb Solution: A tennis ball is dropped on to the floor from a height of 9.8…

A tennis ball is dropped on to the floor from a height of 9.8 m. It rebounds to a height 5.0 m. Ball comes in contact with the floor for 0.2s. The average acceleration during contact is ___________ ms −2^{-2} .

(Given g = 10 ms −2^{-2} )

Step-by-Step Verified Solution

The speed of ball just before collision with ground is

u=2×gH=2×10×9.8=14 m/sec (Downwards) u=\sqrt{2 \times g H}=\sqrt{2 \times 10 \times 9.8}=\underset{\text { (Downwards) }}{14 \mathrm{~m} / \mathrm{sec}}

The speed of ball just after collision is

v=2gh=2×10×5=10 m/sec (Upwards) v=\sqrt{2 g h}=\sqrt{2 \times 10 \times 5}=\underset{\text { (Upwards) }}{10 \mathrm{~m} / \mathrm{sec}}

So, a⃗=Δv⃗Δt\vec{a}=\frac{\Delta \vec{v}}{\Delta t}

=10+140.2=120 m/s2=\frac{10+14}{0.2}=120 \mathrm{~m} / \mathrm{s}^{2}

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