Question

3. Tom standing on the roof of a tall building of 60m height throws a ballat angle of 45 The velocity of the ball is 12 m/s when it leaves tom's hand. Show your complete solution to receive the whole credit for the problem Calculate the following: a-Initial velocity in X and Y direction b-Time to reach to its maximum height. C-Maximum height of the ball with respect to the ground.

Answer 1

- 3 Consider ce is initial Velocity of ball when he projected. Giren U= 12 m/see @@ Ux = U cos 45o =12xt = 652 m/see h Ug = Usinuto = 652 m/ see Time to reach Maximum Light h = Ug & 1 g +² < Vy² = Uyggh W=My /22 {at night paint

VE 5 At night point velocity vzo Vy = Uygt t = toly - Bigid 6. X2 t = 0.87 See H = Gomth = 6om + Uyt- 1 gt² = somt 6x52x0.87 - 1x9.0X607² = 60+ 7.3083-71 H = 63.67m 2 Not defined covere print at which is Velocity calculate." o su=0 - at night paint. 2 Ux=652 au- Un = 6/2 m/ see