Question

A rock is thrown upward with a velocity of 27 meters per second from the top of a 35 meter high cliff on the way back down. When will the Rick be 5 meters from ground level? Round your answer to two decimal places

Answer 1

Given data

*The given veloicty of the rock is u = 27 m/s

*The given height of the cliff is h = 35 m

*The given height of the Rick from the ground level is s = 5 m

*The value of the acceleration doue to gravity is g = 9.8 m/s^2

The formukla for the time taken by the Rick at 5 meters from the ground level is given as

$s=h+ut+\frac{1}{2}at^2$

Substitute the known values in the above expression as

$\begin{gathered} 5=(35)+(27)(t)+\frac{1}{2}(9.8)t^2 \\ 4.9t^2-27t-30=0 \\ t=6.45\text{ s} \end{gathered}$

Hence, the time taken by the Ric at 5 meters above the ground lel is t = 6.45 s