Answer to Question #135369 in Classical Mechanics for Akshaya

Question

A person standing on the edge of a high cliff throws a rock straight up. The rock misses the edge of the cliff as it fall back to earth in 7s. The rock attains maximum height at t = 1.5 s. What is the height of the cliff? What is the total distance travelled?

Answer 1

As per the question,

Total time of flight of the rock $(t)=7 sec$

Taken taken by the rock to the reach the maximum height $(t_1)=1.5 sec$

Let the height of the cliff is $h,$

Taken by rock to reach at the bottom from the top of the cliff $(t_2)=7-1.5 = 5.5 sec$

So,

$h= ut + \dfrac{gt_2^2}{2}$

Now, substituting the values in the above equation,

$\Rightarrow h =0+\dfrac{9.8\times 5.5^2}{2}$

$\Rightarrow h=148.225 m$

When the rock will reach to the maximum height, then it's velocity will become zero, let that is represented by $(v).$

Now, $v=u-gt$

Now, substituting the values in the above,

$0=u-9.8\times 1.5 m/sec$

$\Rightarrow u = 14.7 m/sec$

let the distance traveled by stone in 1.5 sec is represented by d,

$v^2=u^2-2gd$

$\Rightarrow d = \dfrac{u^2-v^2}{2g}$

$\Rightarrow d = \dfrac{u^2}{2g}$

$\Rightarrow d = \dfrac{14.7^2}{2\times 9.8}$

$\Rightarrow d = 11.025m$

Hence, total distance traveled by stone $(D)=d+h$

Substituting the values in the above,

$D=(11.025+148.225)m$

$\Rightarrow D=159.25m$