Question

Two bodies are thrown vertically update with their initial speeds in ratio 2:3.Thee ratio of maximum heights reached by them and thr ratio of their time taken by them to return back to ground respectively are.

Answer 1

hello friend..!!

it is given in the question that the ratio of  the initial speed of the two bodies,

$\frac{u1}{u2}$ = 2/3 

we know that the maximum height when a boly is throwed upward is 

u² / 2g

therefore,

$\frac{h1}{h2}$ = $\frac{u1 ^{2}/2g }{u2 ^{2}/2g }$ 

implies, $\frac{h1}{h2} = \frac{4}{9}$ 

therefore the ratio of maximum height is 4/9 .

the time taken to reach the maximum height is u / g 

therefore$\frac{t1}{t2} = \frac{u1/g}{u2/g}$

therefore $\frac{t1}{t2} = \frac{2}{3}$

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hope it is useful..!!

Answer 2

Explanation:

Initial velocity u = g×t = 9.8×6 = 58.8 m/s

distance S1 travelled in first second is calculated using " S = u×t - (1/2)×g×t2 ", using u = 58.8 m/s, and t = 1 s;

S1 = 58.8×1 -(1/2)×9.8×1×= 53.9 m

Since the paticle reaches maximum height in 6 s, at 7th second travel, its initial velocity is zero.

hence distance S7 travelled in 7th second = (1/2)×9.8×1×1 = 4.9 m

hence the ratio S1 / S7 = 53.9 / 4.9 = 11