a body is dropped from certain height. calculate the time taken to strike the ground
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Answered by
0
S = ut + 1/2gt^2
Since u = 0 we get
S = 1/2gt^2
S=1/2 * 9.8 t^2
4.9t^2
Therefore s = 4.9 t^2
Where s is the height
Since u = 0 we get
S = 1/2gt^2
S=1/2 * 9.8 t^2
4.9t^2
Therefore s = 4.9 t^2
Where s is the height
Answered by
1
s=ut+1/2*a*t²
u=0
s=h
a=9.8 m/s²
h=0*t+1/2*9.8*t²
h=4.9t²
h/4.9=t²
√h/√4.9=t
√(h/4.9)=t
u=0
s=h
a=9.8 m/s²
h=0*t+1/2*9.8*t²
h=4.9t²
h/4.9=t²
√h/√4.9=t
√(h/4.9)=t
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