A ball is thrown vertically upwards from the ground and a boy gazing out the window observes it moving upward past him at 5ms lf g=10 Ms ^2 and the height of the window is 10 m above the ground the ball travels through a height equal to
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♥️HEY MATE♥️
HERE IS YOUR ANSWER
v2−u2=2asv2−u2=2as
u=?v=5m/ss=10ma=−10u=?v=5m/ss=10ma=−10
Therefore 52−u2=2×(−10)×1052−u2=2×(−10)×10
u2=225m/su2=225m/s
Maximumheight=h=u22gMaximumheight=h=u22g
=11.25m=11.25m
Hence a is the correct answer.
HOPE IT HELPS♥️
HERE IS YOUR ANSWER
v2−u2=2asv2−u2=2as
u=?v=5m/ss=10ma=−10u=?v=5m/ss=10ma=−10
Therefore 52−u2=2×(−10)×1052−u2=2×(−10)×10
u2=225m/su2=225m/s
Maximumheight=h=u22gMaximumheight=h=u22g
=11.25m=11.25m
Hence a is the correct answer.
HOPE IT HELPS♥️
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Answered by
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Answer:11.25m
Explanation:There's no need to go by lengthy formulas
All u need to do is use :hmax=hi+hf (note in the question hi is given 10m)
Now hf =u*2/2g =(5)*2/2×10=25/20=1.25
Therefore hmax=10+1.25=11.25m
Just chillax ! Hope it helps you!
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