Date rewarded: October 5th, 2001.

Pawkeet and Mirgle have had their game of

**Deckball**rudely interrupted by a filthy braggart pirate. In an attempt to ruin their fun he steals their deckball, loads it into his cannon and fires it off into the distance.

Assuming that the value of gravity on Neopia is 9.8m/s^{2}, and that the initial launch velocity of the ball is 100 meters per second, and also assuming that the ball lands on a rocky outcrop at exactly the same altitude as the cannon...

**How many seconds is the deckball in the air?**

(rounding down to the nearest second)

Prize

Unknown

Unknown

__Click to show/hide solution__

The information in the problem is not enough to solve. If we want to get the answer that TNT gave, we have to assume the given initial velocity is the part in vertical direction. That means:

v

_{y}= 100 m/sAccording to the trajectory of a projectile equations [1], the time for the deckball to be in the air is:

t = 2vsinθ / g = 2v

_{y}/ g = 20.41 swhere θ is the angle of shooting, and g = 9.8m/s

^{2}is the Neopia gravity acceleration. After rounding, the answer is 20 seconds.

References:

[1] Trajectory of a projectile

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