Q75P - 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  

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.

$$\begin{gathered} t_1=\frac{10}{2} \\ =5\mathrm{s} \\ \frac{5}{{\mathrm{t}}_2}=3 \\ {\mathrm{t}}_2=\frac{5}{3}\mathrm{s} \end{gathered}$$ 

Flight time for slower stone is,

$$\begin{gathered} =2\times \frac{5}{3} \\ =3.33 s \end{gathered}$$

Step 2: Calculation for the height

Lets slower stone's maximum height

$H=\frac{v^2}{2g}$ 

Faster stone's maximum height

$x=\frac{3v^2}{2g}$ 

 

Now equating we get

$$\begin{gathered} \frac{H}{x}=\frac{v^2}{2g}\times \frac{2g}{9x^2} \\ x=9H \end{gathered}$$