A ball is thrown horizontally from the top of a tower with a velocity of 40 ms-1. Take g = 10 ms-2
a) Find the horizontal and vertical displacement after 1,2,3,4,5 seconds, then the path of         the motion of ball. 
b) If the ball reaches the ground in 4 seconds, find the height of the tower.

Dear student,

For horizontal distance, gravity plays no role.

Horizontal Distance=vtSo, in 1,2,3,4,5 sec the distance will be40m, 80m,120m,160m,200mFor vertical distanceS=ut+0.5gt2For 1sec, s=45mfor 2 secS=80+20=100mFor 3 secS=120+45=165mFor 4 secS=160+80=240mFor 5secS=200+125=325m
The path of the ball will be parabolic

(b) When the ball hits the ground, its final velocity is zero
H=u22gH=40220=80m
 
Hope this information will clear your doubts about topic.

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Regards


 

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