Answer :
Answer:
[tex]h=8.261\ m[/tex]
Explanation:
Given:
- time taken by the rock to hit the hit the ground when thrown up from the cliff, [tex]t=2.35\ s[/tex]
- initial velocity of projection, [tex]u=8\ m.s^{-1}[/tex]
Using eq. of motion:
[tex]v_t^2=u^2-2g.s[/tex]
where:
[tex]s=[/tex] distance travelled upwards to reach the top height
[tex]v_t=[/tex] velocity at the top height during the motion = 0
[tex]0^2=8^2-2\times 9.8\times s[/tex]
[tex]s=3.2653\ m[/tex]
Now the time taken to cover this displacement:
[tex]v_t=u-g.t_t[/tex]
where
[tex]t_t=[/tex] time taken to reach the top
[tex]0=8-9.8\ t_t[/tex]
[tex]t_t=0.816\ s[/tex]
Hence the time for which the rock travels down:
[tex]t_d=t-t_t[/tex]
[tex]t_d=2.35-0.816[/tex]
[tex]t_d\approx1.534\ s[/tex]
Now the displacement of the rock in the downward direction:
[tex]y=v_t.t_d+\frac{1}{2} g.t_d^2[/tex]
where:
[tex]y=[/tex] total downward displacement
[tex]v_t=[/tex] initial velocity for the course of downward motion
[tex]t_d=[/tex] total time taken for the downward motion
[tex]y=0+0.5\times 9.8\times 1.534^2[/tex]
[tex]y=11.526\ m[/tex]
So, the height of the cliff:'
[tex]h=y-s[/tex]
[tex]h=11.526-3.265[/tex]
[tex]h=8.261\ m[/tex]