<p>A rocket, initially at rest on the ground, accelerates straight upward from r

Question

<p>A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration <span>39.2</span>&#160;<img src="http://session.masteringphysics.com/render?units=m%2Fs%5E2" alt="m/s^2" align="middle" />. The acceleration period lasts for time 9<span>.00</span>&#160;<img src="http://session.masteringphysics.com/render?units=s" alt="s" align="middle" /> until the fuel is exhausted. After that, the rocket is in free fall</p>

Explanation / Answer

I assume your question to be as follows,
Q) A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 39.2 m/s2 . The acceleration period lasts for time 9.00 s until the fuel is exhausted. After that, the rocket is in free fall.

Ans)
for intiatial 9 s the equation is :

u1 = 0 m/s

a1 = 39.2 m/s2

v1 = u1 +a1t = 39.2*9 = 352.8 m/s

h1 = u1*t + (.5)a1t2 = (.5)39.2*9*9 = 1587.6 m

for 2nd part *free fall case):

u2 = v1 = 352.8 m/s

v2 = 0 m/s

a2 = -9.8 m/s2

v22 = u22 + 2a2h2

so, h2 = u22/(-2a2) = (352.8)2/((-2)*(-9.8)) = 6350.4 m

So maximum height reached by the rocket = 1587.6+6350.4 = 7938m

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