A body is projected with velocity of 40m/s after 2s it crosses a vertical pole of height 20.4m calculate the angle of projection and horizontal range.
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hyy friend here is ur answer
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u = 40 m/s g = 9.8
m/s
At t1 = 2 sec y =
20.4 m
y = u sinФ t - g t² /2
20.4 = 40 sinФ * 2 - g
2² /2
=> sinФ = 1/2
=> Ф = 30°
Horizontal range = R
= u² Sin2Ф / g = 40²
sin60° / 9.8 = 141.39
m
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.
.
.
.
u = 40 m/s g = 9.8
m/s
At t1 = 2 sec y =
20.4 m
y = u sinФ t - g t² /2
20.4 = 40 sinФ * 2 - g
2² /2
=> sinФ = 1/2
=> Ф = 30°
Horizontal range = R
= u² Sin2Ф / g = 40²
sin60° / 9.8 = 141.39
m
...
.
.
.
if u like my answer then mark as brainliast
.
.
.
.
...
and follow me for more
.
.
.
#ar
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