Answer:
(a) 41.75m/s
(b) 4.26s
Explanation:
Let:
Distance, D = 89m
Gravity, [tex]g[/tex] = 9.8 m/[tex]s^{2}[/tex]
Initial Velocity, [tex]u[/tex] = 0m/s
Final Velocity, [tex]v[/tex] = ?
Time Taken, [tex]t[/tex] = ?
With the distance formula, which is
D = [tex]ut[/tex] + [tex]\frac{1}{2} gt^2[/tex]
and by substituting what we already know, we have:
89 = [tex]\frac{1}{2}[/tex]×9.8×[tex]t^{2}[/tex]
With the equation above, we can solve for [tex]t[/tex]:
[tex]t=\sqrt{\frac{89(2)}{9.8}} \\t=\sqrt{\frac{178}{9.8} } \\t=\sqrt{18.16} \\t=4.26 seconds[/tex]
Now that we have solved [tex]t[/tex], we can use the following velocity formula to solve for [tex]v[/tex]:
[tex]v = u + at[/tex], where [tex]a[/tex] is also equals to [tex]g[/tex], so we have
[tex]v = u + gt[/tex]
By substituting [tex]u = 0[/tex], [tex]g = 9.8[/tex], and [tex]t = 4.26[/tex],
We have:
[tex]v = 0 + 9.8(4.26)\\v = 41.75m/s[/tex]
