Question

a ball is dropped from rest at a height of 89m above the ground. (a)what is it's speed just before it hits the ground? (b) how long does it take to reach the ground?​

Answer 1

(a) 41.75m/s

(b) 4.26s

Step-by-step explanation:

Let:

Distance, D = 89m

Gravity,
`g` = 9.8 m/
`s^(2)`

Initial Velocity,
`u` = 0m/s

Final Velocity,
`v` = ?

Time Taken,
`t` = ?

With the distance formula, which is

D =
`ut` +
`(1)/(2) gt^2`

and by substituting what we already know, we have:

89 =
`(1)/(2)`×9.8×
`t^(2)`

With the equation above, we can solve for
`t`:

`t=\sqrt{(89(2))/(9.8)} \\t=\sqrt{(178)/(9.8) } \\t=√(18.16) \\t=4.26 seconds`

Now that we have solved
`t`, we can use the following velocity formula to solve for
`v`:

`v = u + at`, where
`a` is also equals to
`g`, so we have

`v = u + gt`

By substituting
`u = 0`,
`g = 9.8`, and
`t = 4.26`,

We have:

`v = 0 + 9.8(4.26)\\v = 41.75m/s`