A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.36 102 m/s from the top of a cliff 146 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 96.4 m/s at its maximum height of y = 339 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

intial total mechanical enrgy = K.E+P.E =480896+74401.6=555297.6 joules-----1

We know that work done on the system = change in total mechanical energy

total mechanical energy at max height =K.E+P.E=241616.96+172754.4=414371.36 Joules---2

work done by air friction is =(1)-(2) =140926.24 Joules

Total work done done by friction = 2.5*140926.24 =352315.6 joules ---3 (2.5 given by the prob)

so remaining konetic energy E = (1)-(3)=202982 Joules

final speed just before hitting =(2E/m) =88.35 m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.