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

Answer:

(a) 41.75m/s

(b) 4.26s

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 = $ut$ + $\frac{1}{2} gt^2$

and by substituting what we already know, we have:

89 = $\frac{1}{2}$×9.8×$t^{2}$

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

$t=\sqrt{\frac{89(2)}{9.8}} \\t=\sqrt{\frac{178}{9.8} } \\t=\sqrt{18.16} \\t=4.26 seconds$

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:

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