A ball is thrown vertically down with velocity of $5 m / s$ from the top of a tower. With what velocity should another ball be thrown vertically down after $2$ seconds from the top of the same tower, so that it can hit the first ball in a further $2$ seconds (Take $g = 10 m / s^{2}$ )

40 m / s

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55 m / s

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15 m / s

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25 m / s

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Solution

The correct option is A $40 m / s$
For ball 1
u=5 m/s
t=4 secs
For ball 2
u=u m/s
t=2 secs
For collision equating heights
h= $u t + \frac{1}{2} g t^{2}$
$5 \times 4 + \frac{1}{2} 10 \times 4^{2}$ = $u \times 2 + \frac{1}{2} 10 \times 2^{2}$
u=40 m/s

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If a ball is thrown vertically upwards with a velocity of 40 $\frac{m}{s}$ , assuming $(g = 10 \frac{m}{s^{2}})$ then what will be the velocity of the ball after two seconds?

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A ball is thrown vertically down with velocity of 5m/s from the top of a tower. With what velocity should another ball be thrown vertically down after 2 seconds from the top of the same tower, so that it can hit the first ball in a further 2 seconds (Take g=10 m/s2)

The correct option is A 40 m/sFor ball 1 u=5 m/st=4 secsFor ball 2u=u m/st=2 secsFor collision equating heightsh=ut+12gt25×4+1210×42 = u×2+1210×22u=40 m/s

A ball is thrown vertically down with velocity of $5 m / s$ from the top of a tower. With what velocity should another ball be thrown vertically down after $2$ seconds from the top of the same tower, so that it can hit the first ball in a further $2$ seconds (Take $g = 10 m / s^{2}$ )

The correct option is A $40 m / s$
For ball 1
u=5 m/s
t=4 secs
For ball 2
u=u m/s
t=2 secs
For collision equating heights
h= $u t + \frac{1}{2} g t^{2}$
$5 \times 4 + \frac{1}{2} 10 \times 4^{2}$ = $u \times 2 + \frac{1}{2} 10 \times 2^{2}$
u=40 m/s