Question

A gourd is thrown upward off a cliff with a velocity of 5.5m/s. If it lands a total of 3.7s later. How high is the cliff?? How high was it thrown (peak height)? With what velocity does it smash into the ground below?

Answer 1

Given data

*The given initial velocity is u = 5.5 m/s

*The given total time is T = 3.7 s

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

The formula for the time taken by the gourd to reach the highest point of the cliff is given by the equatin of motion as

`v=u-gt`

Here v = 0 m/s is the final velocity of the gourd at the highest point of the cliff

Substitute the known values in the above expression as

`\begin{gathered} 0=5.5-(9.8)t \\ t=0.56\text{ s} \end{gathered}`

The height of the cliff is calculated by the equation of motion as

`\begin{gathered} s_1=ut-(1)/(2)gt^2 \\ =(5.5)(0.56)-(1)/(2)(9.8)(0.56)^2 \\ =1.54\text{ m} \end{gathered}`

Hence, the height of the cliff is s_1 = 1.54 m

The time taken by the gourd to reach the peak height is calculated as

`\begin{gathered} t_1=T-t \\ =3.7-0.56 \\ =3.14\text{ s} \end{gathered}`

The formula for the peak height is given by the equation of motion as

`s_2=(1)/(2)gt^2_1`

Substitute the known values in the above expression as

`\begin{gathered} s_2=(1)/(2)(9.8)(3.14)^2 \\ =48.3\text{ m} \end{gathered}`

Hence, the peak height is s_2 = 48.3 m

The formula for the velocity of the gourd strikes into the ground below is given as

`v=\sqrt[]{2gs_2}`

Substitute the known values in the above expression asv=

`\begin{gathered} v=\sqrt[]{2*9.8*48.3} \\ =30.76\text{ m/s} \end{gathered}`

Hence, the velocity of the gourd smash into the ground is v = 30.76 m/s