A rocket with mass 3950 kg is fired from the ground at an angle of elevation of

Question

A rocket with mass 3950 kg is fired from the ground at an angle of elevation of 60°. The motor creates a force (thrust) on the rocket of 6.2 104 N at a constant angle of 60° to the horizontal for 45 s and then cuts out. As a rough approximation, ignore the mass of fuel consumed and neglect forces from the air. Calculate
(a) the altitude of the rocket at motor cutout and
__________ m
(b) the total horizontal distance from firing point to eventual impact with the ground (assumed to be level).
_____________ m

Explanation / Answer

motor creates a thrust force 6.2 X 104 N at an angle of 60 degrees.
horizontal component = Fx = F cos60 = 31 kN

vertical component Fy = F sin60 = 53.7 kN

ax =31kN / 3950 = 7.85 m/s2

ay = 53.7 kN /3950 =13.6 m/s2

a) distance traveled in vertical direction

s = ut + 1/2(at2 )

Sy = 0(45') + 1/2(13.6 - g)(45)2= 1403.67 m Ans

b) horizontal distance

s = ut + 1/2(at2 ) = 0 X 45 + 1/2(7.85)(45)2= 7948.125 m   Ans

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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