Question

A rocket is fired into the air at an angle of 25 degrees above the horizontal with a velocity of 300 m/s. Using conservation of energy, determin:

a. The maximum height the rocket will attain

b. The work done in order to get the rocket to this height

Answer 1

(a) Solution :  Using energy conservation

We know,

Initial Energy of system = Final Energy of System

K.E. -> Kinetic Energy

P.E. -> Potential Energy

(K.E.)_initial + (P.E.)_initial = (K.E.)_final + (P.E.)_final

M V² / 2 + 0 = 0 + M g H

here, V = V sin25⁰

(resolving velocity vector into components, and considering only vertical component)

H - Max. height

(At max. height rocket has max. P.E, so K.E. is zero)

M(300sin25⁰)² = MX10XH ( Let, g=10m/s² , sin25⁰ = 0.422 )

H = 803.72 m   Answer.

(b) Solution.

Work done in order to get rocket to this height = change in Potential Energy of rocket

= final Potential energy - initial Potential energy

= M g H - 0

= M X 10 X 803.72

= 8037.2 X M joule   Answer.

(since I have taken ground as reference so potential energy at ground is zero. i.e. initial Potential energy is zero)

mass of rocket, M is not given in Question.