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`        A man throws balls with the same speed vertically upwards one after the other at an interval of 2seconds.What should be the speed of throw so that more than two balls are in the sky at any time?`
7 years ago

69 Points
```										Dear chakrala mounica
the first has to be above the ground when the second ball is thrown from the ground

so the first ball has to be above from the ground for 2 seonds.

s=ut-1/2*gt^2        t=2  and s=0 when first ball reaches the ground

0= 2*u- 1/2*g*4

u=g m/s        take g=9.8 m/s^2
u= 9.8 m/s

speed has to be more than 9.8 m/s

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```
7 years ago
Raunak Chaudhury
11 Points
```										S=ut-1/2gt×t S=2u-19.6v^2=u^2-2aS0^2=u^2-2×9.8×(2u-19.6)0=u^2- 39.2u - (19.6)^20=(u- 19.6)^2u=19.6 m/secMore than 2 balls speed is greater than 19.6 m/sec
```
6 months ago
sahil
19 Points
```										more than 19.6 m/s Let the required speed of through be u m/s. The time taken to reach maximum height t=u/g ; For two balls to remain in air at any time, t must be greater than 2. ; ∴ u/g >2 ⇒ u > 2(g) ; u > 19.6 m/s
```
4 months ago
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