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Answer :
- Initial velocity=u=12.5m/s
- Distance=s=38.4m
- Time be t
[tex]\\ \rm\rightarrowtail s=ut+\dfrac{1}{2}gt^2[/tex]
[tex]\\ \rm\rightarrowtail 38.4=12.5t+4.9t^2[/tex]
On solving further we get
- t=1.8s
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A girl on a bridge throws a rock straight down at -12.5 m/s.the time required to hit the ground 38.4 m below is 4.35 s.
Given:
Speed, u = -12.5 m/s
Distance, d = 38.4 m/s
There are three equations of motion present. For the given question, we will use the second equation of motion.
The second equation of motion helps in the determination of velocity, time, and distance. For the given question, we have initial velocity, and distance, to find time we can use the second equation of motion.
From the second equation of motion:
[tex]s = ut + 1/2 at^2\\38.4 = -12.5t - 9.8/2* t^2\\\4.9t^2 - 12.5 t -38.4 = 0[/tex]
Solving the equation, the required time is:
[tex]t = 4.35 s[/tex]
Hence, the time required to hit the ground 38.4 m below is 4.35 s.
To learn more about time, here:
https://brainly.com/question/82806
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