with what speed must a ball be thrown wertical up in order to raise to a maximum height of 45m? and for how long will it be in air?
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considering the accelertion due to gravity to be 10m/s²,
usng the displacement formula s=ut-1/2gt²,
45=ut-5t²---------------(1)
at maximum height of 45m, the final velocity is zero.
so, v=u+at
0=u-10t
∴u=10t-------------------------(2)
substitute eqn(2) in (1), we get
45=10t²-5t²
45=5t²
t²=9 =>t=3s
it stays in the air for three seconds//
so, in eqn(2), put t=3, we get,
u=30m/s
thus, the initial velocity is 30m/s.
hope it helps..
usng the displacement formula s=ut-1/2gt²,
45=ut-5t²---------------(1)
at maximum height of 45m, the final velocity is zero.
so, v=u+at
0=u-10t
∴u=10t-------------------------(2)
substitute eqn(2) in (1), we get
45=10t²-5t²
45=5t²
t²=9 =>t=3s
it stays in the air for three seconds//
so, in eqn(2), put t=3, we get,
u=30m/s
thus, the initial velocity is 30m/s.
hope it helps..
manopollachi:
could u pls mark it the brainliest? please....
u had written gt² isn't it st²??
NO! the symbol for accleratin due to gravity is g!
okay... thnx
but there is no option for marking brainliest.......
when a tab appears near thanks, just click it..
sry i didnt understand.....
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