Answer:
a) time taken = 1.427 s
b) final speed before hitting the ground = 14 m/s
Explanation:
height of cliff = 10 m
acceleration due to gravity = 9.81 [tex]m/s^{2}[/tex]
time taken for stone to reach bottom of cliff = ?
velocity of stone just before hitting the ground = ?
To find the time taken, we will use Newton's equation of motion
using
[tex]v^{2} = u^{2} + 2as[/tex]
where
v = final speed of stone before hitting the ground
u = initial speed of the stone (u = 0 since the stone started falling from rest)
a = acceleration due to gravity = 9.81 [tex]m/s^{2}[/tex]
s = height of the cliff = 10 m
Solving, we have
[tex]v^{2} = 0^{2} + 2(9.81 * 10)[/tex]
[tex]v^{2} = 2*98.1[/tex]
[tex]v^{2} = 196.2[/tex]
[tex]v=\sqrt{196.2} = 14 m/s[/tex]
From the first equation of motion,
[tex]v = u + at[/tex]
where t = time taken for stone to reach bottom of cliff
imputing values, we have
[tex]14 = 0 + (9.81 * t)[/tex]
[tex]t = \frac{14}{9.81} = 1.427 s[/tex]
