Question

A stone of mass m is thrown vertically upwards. Another stone of mass 2m is thrown at an angle theta with vertical. Both of them stay in air for same period of time . The heights attained by the two are in the ratio

Answer 1

Dear Student,
Let stone 1 of mass m is thrown vertically upwards with a velocity v1. The time to reach maximum height will be given by:
$$\begin{gathered} v_f=0=v_1-gt \\ Thus,t=\frac{v_1}{g} \\ Maximumheightattained=h_1 \\ 0={v_1}^2-2g{h}_1 \\ {h}_1=\frac{{v_1}^2}{2g} \\ Timetogoupandreturntotheground=timeofflight=2t=\frac{2v_1}{g} \end{gathered}$$
The stone 2 is thrown at an angle $\theta$ with the vertical i.e $90-\theta$ with respect to the horizontal with a velocity v₂.
It will follow a projectile path and its time of flight = $\frac{2v_2\sin \left(90-\theta \right)}{g}=\frac{2v_2\cos \left(\theta \right)}{g}$
Given, the time of flight of both the stones are same. Thus,
$$\begin{gathered} \frac{2v_1}{g}=\frac{2v_2\cos \left(\theta \right)}{g} \\ Thus,v_1=v_2\cos \left(\theta \right) \end{gathered}$$
The maximum height attained by the stone 2 = h₂
$$\begin{gathered} 0={\left(v_2\cos \left(\theta \right)\right)}^2-2g{h}_2 \\ {h}_2=\frac{{\left(v_2\cos \left(\theta \right)\right)}^2}{2g} \end{gathered}$$
Thus,
$\frac{h_1}{h_2}=\frac{{v_1}^2}{2g}\times \frac{2g}{{\displaystyle {\left(v_2\cos \left(\theta \right)\right)}^2}}=\frac{{\left(v_2\cos \left(\theta \right)\right)}^2}{{\left(v_2\cos \left(\theta \right)\right)}^2}=1$
So, the height attained by both the stones will be the same.
Regards