A body projected vertically upwards with a velocity 49m/s.find the maximum height,time of ascend and time of descent and velocity when it is a height of 60m
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initially velocity = 49m/s.
height =[initially velocity ^2/2g]
=[49×49/2×20]
=120.05m
time of ascend=time of descend.
time of ascend = u/g
=49/10
=4.9second
therefore,time of descend is also 4.9 second.
velocity at 60m:
v^2-u^2=2ah
v^2=u^2-2gh
=49×49-2×10×60
=1201
v =34.65m/s
height =[initially velocity ^2/2g]
=[49×49/2×20]
=120.05m
time of ascend=time of descend.
time of ascend = u/g
=49/10
=4.9second
therefore,time of descend is also 4.9 second.
velocity at 60m:
v^2-u^2=2ah
v^2=u^2-2gh
=49×49-2×10×60
=1201
v =34.65m/s
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