Q.1 a ball is thrown vertically upwards with a velocity of 72km/h from the top of multistory building.the height of the point from where the ball is thrown is 25m from the ground.
(a)how high will the ball rise?
(b)how long will it be before the ball hits the ground?
{take g=10ms–²}
Answers
Answered by
0
Answer:
Time to reach maximum height can be obtained from v=u+at
0=20+(−10)t
t=2s
s=ut+0.5at
2
=20(2)+0.5(−10)(2)
2
=20m
Thus, total distance for maximum height is 45 m
s=ut+0.5at
2
45=0+0.5(10)(t
′
)
2
t
′
=3s
Total time= 3+2= 5s
Answered by
0
Explanation:
let h=25+x
x=ht of building
v2=u2+2as
on eqating, h=5m
v=u+at
t=2s
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