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LenKa [72]
3 years ago
6

A cannon fires a shell straight upward; 1.7 s after it is launched, the shell is moving upward with a speed of 17 m/s. Assuming

air resistance is negligible,a.find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch
(b) 5.5 s after the launch
Physics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

a) 0.323m/s b) 54.278 m/s

Explanation:

The shell is moving upward with a speed of 17m/s

1.7s after it was launch

Using equation of motion on a straight line,

Vfinal = Vintial + acceleration due to gravity *time

Where V represent the speed(velocity)

Vinital = V(at 1.7s) - 9.81* 1.7

Vinital = 17 - 16.677

a) Vinital = 0.323m/s

b) using the same equation,

V(after 5.5s) = Vinital + 9.81*5.5s

Vafter5.5s = 0.323 + 53.955

V after 5.5s = 54.278m/s

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This is a problem of conservation of momentum

Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s

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v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s


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