a ball is thrown from the top of a tower with an intial velocity of 10 ms -1 at an angle of 30 degree with the horizontal if it hits the ground at a distance of 17.3 m from the base of the tower calculate the height of the tower given g = 10ms -2

all the terms used have their usual meanings according to NCERTtextbookIn the Y direction:u = -10 sin 30 = -5 m/sa = 9.8 (10 approx)suppose the height is = htime taken to reach the ground:h = ut+ 0.5at2h = -5t+ 5t2  ........(1)In the x direction!ux = 10 cos 30 = 53 = 8.66m/sa = 0 m/s2s = 17.3 mso the time=  17.3/8.6 = 2 sec put 2 sec in eq 2 h = -5t+ 5t2h = -5(2)+ 5(2)2h = -10+ 20h =10 m 

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