You launch a model rocket from ground level. It moves directly upward with a con

Question

You launch a model rocket from ground level. It moves directly upward with a constant acceleration of 75.0 m/s^2 for 1.45 seconds, at which point it runs out of fuel. Assuming air resistance on the rocket is negligible, what is the maximum altitude (above the ground) achieved by the rocket? Can you find the distance the rocket travels while its engine is on with the given information? Can you then find the distance from the point where the engines run out of fuel to the highest point? What is the final velocity after the initial acceleration period? How can you use this to find the distance traveled while the engines are off? What is the acceleration when the engines are off? What is the rocket's velocity, momentarily, when it reaches the maximum altitude?

Explanation / Answer

speed of the rocket after 1.45 s

v = a*t

= 75*1.45

= 108.75 m/s

height reached by the rocket when fuel runs out, h = 0.5*a*t^2

= 0.5*75*1.45^2

= 78.84 m

so, maximum height reached above the ground, Hmax = h + v^2/(2*g)

= 78.84 + 108.75^2/(2*9.8)

= 682.2 m <<<<<<<<--------------------Answer

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

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