Answers for Part A) 7.0s Part B) x=670 ft Part C) y=110 ft I just want to know h

Question

Answers for Part A) 7.0s Part B) x=670 ft Part C) y=110 ft
I just want to know how to get these answers A U.N. starvation-relief plane dives at angle 53° with the vertical and releases a parcel at an altitude of 2400 ft. The parcel strikes the level ground 5 sec after release. Neglect air resistance (and assume the parcel is shock-proofed). (a) What was the speed of the plane at the instant of release? (b) How far did the parcel travel horizontally during its flight? (c) What are the horizontal and vertical components of the parcel's velocity just before it strikes the ground?

Explanation / Answer

Here ,

theta = 53 degree

t = 5 s

height , t = 2400 feet

a) let the speed of the plane at the instant is u

2400 = u * sin(53) * 5 + 0.5 * 32.17 * 5^2

solving for u

u = 500.32 feet/s

the speed of the plane at the time of release is 500.32 feet/s

b)

horizontal distance = 500.32 * cos(53) * 5 feet

horizontal distance = 1505 feet

c)

at the ground

horizontal velocity = 500.32 * cos(53)

horizontal velocity = 301 feet/s

vertical velocity = 32.17 * 5 + 500.32 * sin(53)

vertical velocity = 560.4 feet/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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