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

6.45 s

Step-by-step explanation:

df = final position = 5 m a = accel of gravity = -9.81 m/s^2

do = original position = 35 m vo = original velocity = 27m/s

df = do + vot + 1/2 at^2

5 = 35 + 27t - (1/2)9.81t^2

-4.905 t^2+ 27t +30 = 0

Use quadratic formula a = -4.905 b = 27 c = 30

to find t = 6.45 s

(Ignore the negative value found with the Quadratic Formula)

Answer 2

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+(1)/(2)at^2`

Substitute the known values in the above expression as

`\begin{gathered} 5=(35)+(27)(t)+(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