Q75 P - University Physics with Modern Physics

Question

Two stones are thrown vertically upward from the ground, one with[HA1]  three times the initial speed of the other. (a) If the faster stone takes 10 s to return to the ground, how long will it take the slower stone to return? (b) If the slower stone reaches a maximum height of H, how high (in terms of H) will the faster stone go? Assume free fall.

Step-by-Step Solution

Answer

a) The time taken by the slower stone is 3.33 s.

b) The faster stone will go 9H. 

Step1: Identification of given data

The slower time taken is 9 s.

Step 2: Calculation for the time

Let elapsed time to peak point for faster stone be \({t_1}\) 

The velocity of the faster stone

\({t_1} = \frac{{3v}}{g}\)

Let elapsed time to peak point for slower stone be 

The velocity of the slower stone

\({t_2} = \frac{v}{g}\) 

Let's find the ratio 

\(\begin{array}{l}\frac{{{t_1}}}{{{t_2}}} = \frac{{3v}}{g} \times \frac{g}{v}\\\frac{{{t_1}}}{{{t_2}}} = 3\end{array}\) 

Since the movement is symmetric, the time to reach 

The maximum point is equal to half of the duration of the flight.

\(\begin{array}{c}{t_1} = \frac{{10}}{2}\\ = 5\,{\rm{s}}\\\frac{5}{{{t_2}}} = 3\\{t_2} = \frac{5}{3}\,{\rm{s}}\end{array}\) 

Flight time for slower stone is,

\(\begin{array}{l} = 2 \times \frac{5}{3}\\ = 3.33\,{\rm{s}}\end{array}\)

Step 3: Calculation for the height

Lets slower stone's maximum height

\(H = \frac{{{v^2}}}{{2g}}\) 

Faster stone's maximum height

\(x = \frac{{3{v^2}}}{{2g}}\) 

 

Now equating we get

\(\begin{array}{l}\frac{H}{x} = \frac{{{v^2}}}{{2g}} \times \frac{{2g}}{{9{x^2}}}\\x = 9H\end{array}\)