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Standard XII

Physics

1st Equation of Motion

A ball is dro...

Question

A ball is dropped from the top of a tower. Another ball is thrown up vertically with a velocity of $20$ $m / s$ from the ground level at the same instant and meets the first ball after $1.5$ seconds. Height of the tower is

20 m

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30 m

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40 m

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50 m

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Solution

The correct option is D 30 m
For A: $s_{A} = \frac{1}{2} g t^{2}$
For B: $s_{B} = 20 (t) - \frac{1}{2} g t^{2}$
$s_{A} + s_{B} = 20 t = H$
$t = 1.5 s e c$
$\Rightarrow H = 30 m$

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A ball is dropped from the top of a tower. Another ball is thrown up vertically with a velocity of 20 m/s from the ground level at the same instant and meets the first ball after 1.5 seconds. Height of the tower is

The correct option is D 30 mFor A: sA=12gt2For B: sB=20(t)−12gt2sA+sB=20t=Ht=1.5 sec⇒H=30 m

A ball is dropped from the top of a tower. Another ball is thrown up vertically with a velocity of $20$ $m / s$ from the ground level at the same instant and meets the first ball after $1.5$ seconds. Height of the tower is

The correct option is D 30 m
For A: $s_{A} = \frac{1}{2} g t^{2}$
For B: $s_{B} = 20 (t) - \frac{1}{2} g t^{2}$
$s_{A} + s_{B} = 20 t = H$
$t = 1.5 s e c$
$\Rightarrow H = 30 m$