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A ball is thrown horizontally with a speed of 2p m/s from a tall tower , the tangential acceleration after 2 seconds is (take g=10 m/s2)
Hiii Here uy = 0 t = 2 second ay = 10 m/s^2vy = ?vy = uy + atvy = 20 m/sso magitude of tangential acceleration is vector sum of x and y component of velocity so, Vt = sqrt ( 400 + 400)Vt = 20root2 Caution : I have considered horizontal velocity as 20 m/s
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