A stone was drapped off a cliff and hit the ground with a speed of 104 t/s. what is the height of the cilff (use 32 t/s2 for the acceleration due to gravity.)
Answers
Answered by
0
Hii friend,
Given,
velocity => 104 m/s
acceleration => 32 m/s²
height=?
Formula:-
v²=u² +2aS
S = v²/2a
S = 104×104/2×32
S = 10816 /64
S =169 m.
The distance of the cliff is 169m.
Hope this helps you a little!!
Given,
velocity => 104 m/s
acceleration => 32 m/s²
height=?
Formula:-
v²=u² +2aS
S = v²/2a
S = 104×104/2×32
S = 10816 /64
S =169 m.
The distance of the cliff is 169m.
Hope this helps you a little!!
Answered by
3
Hey.
Here is your answer.
as for free falling body v=√2gs
so, 104=√2×32×s
so, 104^2=2×32×s
so, 104×104/2×32= s
so, s =169 ft.
Thanks.
Here is your answer.
as for free falling body v=√2gs
so, 104=√2×32×s
so, 104^2=2×32×s
so, 104×104/2×32= s
so, s =169 ft.
Thanks.
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