Atest rocket is fired straight up from rest. the net acceleration is 20 m/s2 upward and continues for 4.0 seconds, at which time the rocket engines cease firing. what maximum elevation does the rocket reach?

Maximum elevation, h = 160 meters

Explanation:

Initially, the rocket is at rest, u = 0

Acceleration of the rocket,

Time, t = 4 s

We need to find the maximum elevation reached by the rocket. It can be calculated using second equation of motion as :

h = 160 meters

So, the maximum elevation reached by the rocket is 160 meters. Hence, this is the required solution.

 486.5 m  

Explanation:

Initial velocity is zero as the rocket is fired from rest. u = 0.

Displacement of the rocket during this time:

s = ut +0.5 at²

s = 0+0.5 ×20×4²

s = 160 m

The final velocity at the end of 4 s is:

v = u + at

v = 0 + (20)(4)

v = 80 m/s

This will become the initial velocity for the next half of the motion.

At the maximum elevation, velocity is zero. v = 0

Acceleration due to gravity always acts downwards.

Thus, the maximum elevation that test rocket would reach is:

326.5 m + 160 m = 486.5 m

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.