Question 186: Laws of Motion

Question

A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in contact with floor for 1 s . The average acceleration during contact is (acceleration due to gravity $=10 \mathrm{~m} / \mathrm{s}^2$ )

Accepted Answer

Velocity just before hitting the ground is
$$ \mathrm{v}_1=\sqrt{2 \mathrm{gh}_1}=\sqrt{2 	imes 10 	imes 20}=20 \mathrm{~m} / \mathrm{s} $$
Velocity just after hitting the ground is
$$ \mathrm{v}_2=\sqrt{2 \mathrm{gh}_2}=\sqrt{2 	imes 10 	imes 5}=10 \mathrm{~m} / \mathrm{s} $$
After hitting the ground ball goes in opposite direction, $	herefore \mathrm{v}_2$ will be $-10 \mathrm{~m} / \mathrm{s}$
$$ a_{a v}=\frac{\Delta v}{\Delta t}=\frac{20-(-10)}{1}=30 \mathrm{~m} / \mathrm{s}^2 $$

Answer

$30 \mathrm{~m} / \mathrm{s}^2$

Answer

$20 \mathrm{~m} / \mathrm{s}^2$

Answer

$40 \mathrm{~m} / \mathrm{s}^2$

Answer

$35 \mathrm{~m} / \mathrm{s}^2$

Laws of Motion

A ball is dropped on the floor from a height of 20 m . It rebounds to a height of 5 m . Ball remains in contact with floor for 1 s . The average acceleration during contact is (acceleration due to gravity =10 \mathrm{~m} / \mathrm{s}^2 )

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