# A bullet fired straight up from the surface of the moon would reach a height of s = 832*t - 2.6*t^2 m after t sec. On earth, in the absence of air, its height would be s = 832*t - 16*t^2 m after...

A bullet fired straight up from the surface of the moon would reach a height of s = 832*t - 2.6*t^2 m after t sec. On earth, in the absence of air, its height would be s = 832*t - 16*t^2 m after t second. How long will bullet be aloft in each case?

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### 1 Answer

The distance traveled by a bullet fired at 832 m/s after t seconds on the Moon is given by: s = 832*t - 2.6*t^2 m

The bullet stops when the speed is zero. When speed is 0, ds/dt = 0

=> 832 - 5.2t = 0

=> t = 832/5.2

=> t = 160

The time the bullet is aloft on the Moon is 160 s

For Earth, the distance traveled by a bullet fired at 832 m/s after t s is given by s = 832*t - 16*t^2 m

It stops when ds/dt = 0

=> 832 - 32t = 0

=> t = 26 s

**The bullet is aloft for 160 s on the Moon and for 26 seconds on the Earth.**