# A ball is allowed to fall from the top of tower of 200m high. At the same instant another ball is thrown vertically upwards from the bottom of the tower with a velocity of 40m/s. When and where the two balls meet?

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The position of the ball from the top of the tower after time t is

h1 = ut + 1/2 g t² = 0 + 1/2 * g * t²

height above ground of the ball thrown from ground - after time t is

h2 = u t - 1/2 g t² , u = 40 m/sec

h1 + h2 = 200 meters

u t - 1/2 g t² + 1/2 g t² = 200 meters

t = 200 / u = 5 seconds

h2 = height above ground = 200 - 1/2 g t² = 75 meters

h1 = ut + 1/2 g t² = 0 + 1/2 * g * t²

height above ground of the ball thrown from ground - after time t is

h2 = u t - 1/2 g t² , u = 40 m/sec

h1 + h2 = 200 meters

u t - 1/2 g t² + 1/2 g t² = 200 meters

t = 200 / u = 5 seconds

h2 = height above ground = 200 - 1/2 g t² = 75 meters