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1st Equation of Motion

At t=0 ,...

Question

At $t = 0$ , an arrow is fired vertically upwards with a speed of $98 m s^{- 1}$ . A second arrow is fired vertically upwards with the same speed at $t = 5 s$ . Then, (Take $g = 9.8 m s^{- 2}$ )

the arrows will be at the same height above the ground at

t = 12.5 s

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the two arrows will reach back at their starting points at

t = 20 s

and

t = 25 s

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the ratio of the speeds of the first and the second arrows at

t = 20 s

will be

2 : 1

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all are correct.

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Solution

The correct option is B all are correct.
$s = u t + \frac{1}{2} a t^{2} \Rightarrow 98 t - \frac{1}{2} 9.8 t^{2} = 98 (t - 5) - \frac{1}{2} 9.8 {(t - 5)}^{2} \Rightarrow t^{2} - {(t - 5)}^{2} - 100 = 0 \Rightarrow 10 t = 125 t = 12.5 s$
Now, total time for first arrow will be
$s = u t + \frac{1}{2} a t^{2} \Rightarrow 98 t - \frac{1}{2} 9.8 t^{2} = 0 t (t - 20) = 0 t_{1} = 20 s \Rightarrow t_{2} = 25 s$
Now at $t = 20 s$
$v = u + a t v_{1} = 98 & v_{2} = 98 \frac{1}{2} \frac{v_{1}}{v_{2}} = \frac{2}{1}$

Suggest Corrections

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