Question

4. A rocket on a stationary launch pad is accelerated upward at 5 m/s2 for 7 seconds during the launch phase, and then coasts to the top. a. what is the velocity and height at the end of the launch phase? b. what is the velocity and height at the end of the coast phase

Answer 1

(a) For launch phase,

Initial velocity = u₁ = 0 m / s,

Acceleration = a = 5 m / s²,

Time of the phase = t = 7 s.

Hence, final velocity = v₁ = u₁ + at = ( 0 + 5 x 7 ) m / s

or, Final velocity at the end of the launch phase = 35 m / s.

Height at the end of the launch phase is : h₁ = u₁t + at² / 2 = ( 0 + 5 x 7² / 2 ) m = 122.5 m.

(b) For coast phase,

Initial velocity = u₂ = v₁ = 35 m / s,

Deceleration = g = 9.8 m / s², since only gravity is acting on the rocket,

Final velocity = v₂ = 0 m / s, since, at the end of coasting the rocket will momentarily stop.

Hence, final velocity at the end of the coast phase = 0 m / s.

Vertical distance covered during the coast phase, h₂, in m, is given by : v₂² = u₂² - 2gh₂

or, 0 = u₂² - 2gh₂

or, h₂ = u₂² / 2g = 35² / ( 2 x 9.8 )

or, h₂ ~ 62.5.

Hence, height at the end of the coast phase = h₁ + h₂ = 122.5 m + 62.5 m = 185 m.