A 52.0 -kg projectile is fired at an angle of 30.0 A 52.0-kg projectile is fired

Question

A 52.0-kg projectile is fired at an angle of 30.0

A 52.0-kg projectile is fired at an angle of 30.0 degree above the horizontal with an initial speed of 1.42 102 m/s from the top of a cliff 118 m above level ground, where the ground is taken to be y = 0. What is the initial total mechanical energy of the projectile? J Suppose the projectile is traveling 100.6 m/s at its maximum height of y = 328 m. How much work has been done on the projectile by air friction? J What is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is going down as it did when it was going up? m/s

Explanation / Answer

mgh+.5mv^2=ME

52*9.8*118+.5*52*(1.42*10^2)^2=584.4 KJ

b)from work energy principle

work done by air friction=change in energy

Energy at max hight=.5*52*100.6^2+52*9.8*328=430.27KJ

work done by air friction=584.4 KJ-430.27KJ=154.13 KJ

C)work done by air friction GOING UP IS 154.13 KJ

while going down is 1.5*154.13=231.2 KJ

total=385.32 KJ

from work energy peinciple

(584.4-385.32)1000=.5*52*v^2

v=87.5 m/s

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