find how high will it rise and how much time will it take to return to its point of projection if a body is thrown up with a velocity 748.4m/s
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using 1st equation of uniformly accelerated motion to find time taken to return to the point of projection.
v = u + g t
(taking g = 10 m/s²)
0 = 748.4 - 10 x t
10 x t = 748.4
t = 748.4/10
time taken = 74.84 seconds.
it will take 74.84 topmost level and 149.68 seconds to return to the point of projection.
using 2nd equation of uniformly accelerated motion to find the height.
h=u t +1/2 g t²
(taking g = 10 m/s²)
h =0 x 74.84 + 1/2 x 10 x 74.84 x 74.84
h =1 / 2 x 10 x 74.84 x 74.84
h ≈ 28005 metres.
it will reach 28005 metres high if it is thrown up with a velocity of 748.4 m/s
PLEASE MARK IT AS A BRAINLIEST ANSWER.
v = u + g t
(taking g = 10 m/s²)
0 = 748.4 - 10 x t
10 x t = 748.4
t = 748.4/10
time taken = 74.84 seconds.
it will take 74.84 topmost level and 149.68 seconds to return to the point of projection.
using 2nd equation of uniformly accelerated motion to find the height.
h=u t +1/2 g t²
(taking g = 10 m/s²)
h =0 x 74.84 + 1/2 x 10 x 74.84 x 74.84
h =1 / 2 x 10 x 74.84 x 74.84
h ≈ 28005 metres.
it will reach 28005 metres high if it is thrown up with a velocity of 748.4 m/s
PLEASE MARK IT AS A BRAINLIEST ANSWER.
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