A 12.0-{\ m kg} shell is launched at an angle of 55.0 ^\\circ above the horizont

Question

A 12.0-{ m kg} shell is launched at an angle of 55.0 ^circ above the horizontal with an initial speed of 150 { m{ m/s}}. When it is at its highest point, the shell exploded into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Assume that air resistance can be ignored. If the heavier fragment lands back at the same point from which the shell was launched, where will the lighter fragment land and how much energy was released in the explosion?

Explanation / Answer

cos they reach fround at exactly the same time, themomentum in y-direction remain unchanged. so, right after the explosion, they don't have y-component ofvelocity, they only have vx because if one of them go up, the other one will go down, thenthe time they reaching ground will surely not be the same. so, find vx vx = v cos55 = 86 m/s the heavier one will have the same but negative velocity as itget back to the launching point. m vx = 3/4m (-vx) + 1/4mv2 v2 = 7vx = 602.2 m/s find t to reach the highest point v (at highest point) = vsin55 - gt t = 12.538 s the point where the lighter mass will land is: x = vx t + v2 t = 1078.3 + 7550.4 =8628.67 m away from the launching point energy released = E E = Efinal - Einitial       = 1/2 1/4 m 602.22 +1/2 3/4 m 862 - 1/2 m862               (m=12)       = 532873.26 J = 532.9 kJ       = 532873.26 J = 532.9 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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