# Calculate the height of a cliff if it takes 2.35 s for a rock to hit the ground when it is thrown straight up from the cliff

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## Answers

We have the equation of motion , [tex]s =ut+\frac{1}{2}at^2[/tex], where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

In this case s = ?

We have u = 8 m/s, t = 2.35 seconds and a = -9.81 [tex]m/s^2[/tex]

Substituting

[tex]s =8.00*2.35-\frac{1}{2}*9.81*2.35^2\\ \\ s = -8.29 m[/tex]

Here - ve sign means below the level of cliff it is 8.29 meter deep

So height of cliff = 8.29 meter

[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]

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