Question

Two stones are thrown vertically upwards simuntaneously with their initial velocity U1 and U2 respectively. Prove that rhe heights reached by them would be in the ratio u1square:u2sqaure

Answer 1

Maximum height reached by an object thrown with vertically upward velocity u is given by

$h = \frac{ {u}^{2} }{2g}$

Height h1 reached by object with initial velocity u1 is:

$h_1 = \frac{ {(u_1)}^{2} }{2g} = \frac{ {u}^{2}_1}{2g}$

Height h2 reached by object with initial velocity u2 is:

$h_2 = \frac{ {(u_2)}^{2} }{2g} = \frac{ {u}^{2}_2}{2g}$

Ratio of h1 and h2:

$\frac{h_1}{h_2} = \frac{ {u}^{2}_1}{2g} \div \frac{ {u}^{2}_2}{2g} \\ \\ \frac{h_1}{h_2} = \frac{ {u}^{2}_1}{2g} \times \frac{2g}{ {u}^{2}_2}$

$\frac{h_1}{h_2} = \frac{ {u}^{2}_1}{ {u}^{2}_2 }$

Answer 2

Maximum height reached by stone1 is
2*g*h1=v²-u1²
2gh1=-u1²
so, h1=-u1²/2g .............(1)
Maximum height reached by stone2 is
2g2=v²-u2²
2gh2=-u2²
h2=-u2²/2g ............ (2)

Now ratio of the height= h1/h2
= -u1²/2g*2g/-u2²
= -u1²/-u2²
= u1²/u2²
Hence, proved