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A rock is thrown downward from an unknown height above the ground with an initial speed of 21m/s. It strikes the ground 8s later. Determine the initial height of the rock above the ground. The acceleration of gravity is 9m/s/s. Answer in units of m



Answer :

rabinshrestha41
Given:
initial velocity (u) = 21 m/s
time (t) = 8 seconds
acceleration (a) = 9 m/s²
distance (s) = ????????
Now,
you can use the formula, [tex]s=ut+ \frac{1}{2} a t^{2} [/tex]

Now, plug the values in the formula, and you get:

[tex]s=21*8+ \frac{1}{2} *9 * 8^{2} [/tex]

[tex]s=21*8+ \frac{1}{2} *9 * 64[/tex]

[tex]s=21*8+ \frac{1}{2} *576[/tex]

[tex]s=21*8+ 288[/tex]

[tex]s=168+288[/tex]

[tex]s= 456~meters[/tex]

SO THE INITIAL HEIGHT OF THE ROCK WAS 456 METERS.




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