A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise?

5 m

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3.75 m

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2.50 m

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1.25 m

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Solution

The correct option is D $5 m$
A/q Ball takes $1 s$ to reach hieghest point,
hence initial velocity,
$u = a t$
$u = 10 \times 1 = 10 m / s$
$u^{2} = 2 g h$
$⟹ h = \frac{u^{2}}{2 g}$
$⟹ h = 10^{2} / (2 \times 10) = 5 m$

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A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise?

The correct option is D 5mA/q Ball takes 1s to reach hieghest point,hence initial velocity,u=atu=10×1=10m/su2=2gh⟹h=u22g⟹h=102/(2×10)=5m

The correct option is D $5 m$
A/q Ball takes $1 s$ to reach hieghest point,
hence initial velocity,
$u = a t$
$u = 10 \times 1 = 10 m / s$
$u^{2} = 2 g h$
$⟹ h = \frac{u^{2}}{2 g}$
$⟹ h = 10^{2} / (2 \times 10) = 5 m$