Kinematics

JEE Main level, numerical answer, medium

158 of 308

Question 158 of 308

A tennis ball is dropped on to the floor from a height of 9.8m9.8m. It rebounds to a height 5.0m5.0m. Ball comes in contact with the floor for 0.2s0.2s. The average acceleration during contact is ______ ms−2m{s}^{-2}. [Given g=10ms−2g=10m{s}^{-2}]

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Answer: 120

The speed of ball just before collision with ground is vi2=0+2ghi{v}_{i}^{2}=0+2g{h}_{i}
⇒vi=2ghi\Rightarrow {v}_{i}=\sqrt{2g{h}_{i}}
=2×10×9.8=\sqrt{2\times 10\times 9.8}
=14ms−1(Downward)=14m{s}^{-1}\left(Downward\right)
The speed of ball just after collision is 0=vf2−2ghf0={v}_{f}^{2}-2g{h}_{f}
⇒vf=2ghf\Rightarrow {v}_{f}=\sqrt{2g{h}_{f}}
=2×10×5=\sqrt{2\times 10\times 5}
=10ms−1(Upward)=10m{s}^{-1}\left(Upward\right)
Average acceleration of ball is ∣a⃗avg∣=∣Δv⃗Δt∣=vf−(−vi)t=10+140.2=240.2=120ms−2\left|{\vec{a}}_{avg}\right|=\left|\dfrac{\Delta \vec{v}}{\Delta t}\right|=\dfrac{{v}_{f}-\left(-{v}_{i}\right)}{t}=\dfrac{10+14}{0.2}=\dfrac{24}{0.2}=120m{s}^{-2}.

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