a ball is dropped from a height of 49m. find the time interval in which the ball hits the ground (g=9.8m/s^2)
Answers
Answered by
5
Given that ,
U = 0
S = 49 m
a = 9.8 m / s ^2
S = ut + 1 / 2 at ^2
49 = 1/2 * 9.8 * t^2
49 = 4.9 * t ^ 2
t ^ 2 = 49 / 4.9
t ^ 2 = 10
t = √ 10 sec.
U = 0
S = 49 m
a = 9.8 m / s ^2
S = ut + 1 / 2 at ^2
49 = 1/2 * 9.8 * t^2
49 = 4.9 * t ^ 2
t ^ 2 = 49 / 4.9
t ^ 2 = 10
t = √ 10 sec.
mdtabseerpbg3f7:
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Answered by
4
hey here is your answer
u=0
t=?
h=49m
g=9.8m/s^2
using equation of motion
h=ut+1/2gt^2
49=0+1/2*9.8*t^2
t^2=49*2/9.8
t^2=10
t=√10 second
u=0
t=?
h=49m
g=9.8m/s^2
using equation of motion
h=ut+1/2gt^2
49=0+1/2*9.8*t^2
t^2=49*2/9.8
t^2=10
t=√10 second
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