Q2.7-47PE - College Physics (Urone)

Question

a) Calculate the height of a cliff if it takes \({\bf{2}}.{\bf{35}}{\rm{ }}{\bf{s}}.\) for a rock to hit the ground when it is thrown straight up from the cliff with an initial velocity of\({\bf{8}}.{\bf{00}}{\rm{ }}{\bf{m}}/{\bf{s}}\). (b) How long would it take to reach the ground if it is thrown straight down with the same speed?

Step-by-Step Solution

Answer

(a) \(8.30 m\) 

(b) \(0.719\)sec.

Describing the problem

The senior of the above problem is related to that of the free fall of the body

Initially the rock is thrown up. 

Let's calculate the time and distance for the same

Given data for the first case

U=\(8.00 m/s\)

V= \(0\)

g= \( - 9.81 m/{s^2}\)

The unknown factors are 

Height covered by stone going up =\(\;h\) 

Time taken by the stone in going up=\(\;t\) 

Height or distance covered can be calculated by

\(\begin{array}{c}{v^2} - {u^2} = 2gd\\{\left( 0 \right)^2} - {\left( 8 \right)^2} = 2\left( { - 9.81} \right)d\\d = 3.261\,\,m\end{array}\)

Hence the height covered by the stone when thrown up is \(3.261 m\)

The time taken by the stone to reach the top position 

\(\begin{array}{l}v = u + gt\\0 = 8 + \left( { - 9.81} \right)t\\t = 0.815\,\,s\end{array}\)

Now the second half of the motion of the rock is performed from top position to downwards.

In this case the initial velocity will be zero.

The time take for the stone to come in ground will be total time subtracted by the time taken to reach the top position.

Hence time

\(\begin{array}{l} = 2.35 - 0.815 \\ = 1.535{\rm{ }}s\end{array}\)

Hence the time taken to travel the stone from top to ground is \(1.485{\rm{ }}s\) 

The height of the cliff

The height of the cliff can be calculated as

\(\begin{array}{l}h = ut + \frac{1}{2}g{t^2}\\h = \left( 0 \right)\left( {1.535} \right) + \frac{1}{2}\left( {9.81} \right){\left( {1.535} \right)^2}\\h = 11.56\,\,m\end{array}\)

Hence the height of the cliff is 

\(\begin{array}{*{20}{l}}{ =  h - d}\\{ = 11.56 - 3.261}\\{ = 8.30}\end{array}\)

The height of the cliff is \(8.30{\rm{ }}m\) 

Time if the stone is thrown directly downwards

If the stone is directly thrown downwards the initial velocity will be \(8{\rm{ }}m/s\) 

The distance traveled by the stone will be equal to the height of the cliff.

Acceleration will be gravitational acceleration.

Hence the given data 

\(\begin{array}{*{20}{l}}{U = 8{\rm{ }}m/s}\\{H = {\rm{ 8}}{\rm{.30}}}\\{g = 9.81{\rm{ }}m/{s^2}}\end{array}\) 

\(\begin{array}{l}h = ut + \frac{1}{2}g{t^2}\\8.30 = \left( 8 \right)\left( t \right) + \frac{1}{2}\left( {9.81} \right){\left( t \right)^2}\\4.905{t^2} + 8t - 8.30 = 0\end{array}\)

\(\begin{array}{l}t = 0.719\,\,\,\,or\\t =  - 2.35\,\,s\end{array}\)

Because time cannot be negative, hence the value of time if the stone is thrown directly is \(0.719\)sec.