13. 0.51 points I Previous Answers Tiplere 6 P067 My Notes Ask Your Teacher The

Question

13. 0.51 points I Previous Answers Tiplere 6 P067 My Notes Ask Your Teacher The initial kinetic energy imparted to a 0.0270 kg bullet is 1181 J. (a) Assuming it accelerated down a 1.00 m long rifle barrel, estimate the average power delivered to it during the firing. 74 kW (b) Neglecting air resistance, find the range of this projectile when it is fired at an angle such that the range equals the maximum height attained. X km 1.00 Assume that the firing height is negligible and that the bullet lands at the same elevation from which it was fired. Use acceleration equations in the horizontal and vertical directions to find the range of the bullet and its maximum height. The bullet's initial speed can be determined from its initial kinetic energy. eBook

Explanation / Answer

1/2*m*v^2 = 1181
1/2*0.027*v^2 = 1181
v = 295.77 m/s

We know Range of a projectile = Maximum height acheieved
vo^2 * sin(2)/g = vo^2 * sin^2()/2g
sin(2) = sin^2() / 2
4*sin()*cos() = sin^2()
tan() = 4
= 76o

Range, R = vo^2 * sin(2)/g = 295.77^2 * sin(152) / 9.8
R = 4190 m
R = 4.19 km

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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