A body is dropped from a height of 49m. Find the time taken by the body to reach the ground ?
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Answered by
4
s = ut + (1/2*a*t^2)
u = 0 (dropped from a height therefore was stationary in the beginning)
s = 49 m
a = 9.8m/s^2
Therefore,
49 = 0 + (1/2 * 9.8 * t^2)
49 = 4.9 * t^2
10 = t^2
t = √10
u = 0 (dropped from a height therefore was stationary in the beginning)
s = 49 m
a = 9.8m/s^2
Therefore,
49 = 0 + (1/2 * 9.8 * t^2)
49 = 4.9 * t^2
10 = t^2
t = √10
prmkulk1978:
please specify the units while solving physics numericals
Answered by
3
Given :
Height =h =49m
Initial speed= u= 0 m/s as dropped from height.
time =t=?
a=g=9.8m/s²
From the second equation of motion :
s=ut+1/2at²
49=0xt +1/2*9.8*t²
49=4.9 t²
t²=49/4.9
=490/49
t²=10sec
t=√10 sec=3.16 sec
∴ Time taken by the body to reach the ground is 3.16sec
Height =h =49m
Initial speed= u= 0 m/s as dropped from height.
time =t=?
a=g=9.8m/s²
From the second equation of motion :
s=ut+1/2at²
49=0xt +1/2*9.8*t²
49=4.9 t²
t²=49/4.9
=490/49
t²=10sec
t=√10 sec=3.16 sec
∴ Time taken by the body to reach the ground is 3.16sec
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