A man drops a ball downside from the roof of a tower of height 400 meters- At the same time another ball is thrown upside with a velocity 50 meter-sec- from the surface of the tower- then they will meet at which height from the surface of the tower

Let both balls meet at point P after time t. The distance travelled by ball A, h_1 = 1/2gt^2 The distance travelled by ball B, h_2 = ut 1/2gt^2 h_1 + h_2 = 4 ...

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Standard XII

Physics

General Gravity Problem

A man drops a...

Question

A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity $50 \frac{m e t e r}{s e c}$ . from the surface of the tower, then they will meet at which height from the surface of the tower

100 meters

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320 meters

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80 meters

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200 meters

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Solution

The correct option is A 100 meters
Let both balls meet at point P after time t.
The distance travelled by ball A, $h_{1} = \frac{1}{2} g t^{2}$
The distance travelled by ball B, $h_{2} = u t - \frac{1}{2} g t^{2}$
$h_{1} + h_{2} = 400 m \Rightarrow u t = 400, t = \frac{400}{50} = 8 s e c$
$∴$ $h_{1} = 320 m$ and $h_{2} = 80 m$

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A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50metersec. from the surface of the tower, then they will meet at which height from the surface of the tower

The correct option is A 100 meters Let both balls meet at point P after time t. The distance travelled by ball A, h1=12gt2 The distance travelled by ball B, h2=ut−12gt2 h1+h2=400m⇒ut=400,t=40050=8sec ∴ h1=320m and h2=80m

A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity $50 \frac{m e t e r}{s e c}$ . from the surface of the tower, then they will meet at which height from the surface of the tower