A rocket blasts off vertically from rest on the launch pad with an upward accele

Question

A rocket blasts off vertically from rest on the launch pad with an upward acceleration of 2.70m/s2 . At 10.0s after blastoff, the engines suddenly fail, which means that the force they produce instantly stops.

How high above the launch pad will the rocket eventually go?

hmax= ____________m

Find the magnitude of the rocket's acceleration at its highest point.

avertex= ____________m/s^2

How long after it was launched will the rocket fall back to the launch pad?

t=___________________s

How fast will it be moving when it does so?

v=_____________________m/s

Explanation / Answer

a=2.7 m/s^2==>t=10 s==>v=2.7*10=27 m/s

ht=1/2*27*10^2=135 m==>ht after engine failure=27^2/(2*9.91)=37.156 m

total ht=37.156+135=172.156 m

a) max ht=172.156 m

b) after engine failure a=-9.81 m/s^2

c) t=(-27-sqrt(27^2+1/2*4*9.81*135))/-9.81=8.677 seconds

total time=10+8.677=18.677 s

answer 18.7 seconds

d) sqrt(2*9.81*172.156)=58.118 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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