Lesly24
Answered

Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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

Sagot :

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.