Question

1.A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30 m/s How long did the burn phase last 2.What is the max height the rocket will reach

Answer 1

1. t = 3.27 seconds

2. y = 147.3 m

Step-by-step explanation:

Newton's Laws of Motions.

y = v₁t + 1/2 at²

a = (v₂-v₁)/t

where

y = the vertical distance travelled

v₁ = the initial velocity

v₂ = the final velocity

t = the time

a = the acceleration

final velocity is equal to 0.

So, v₂ = 0.

a = (v₂-v₁)/t

a = (0-30)/t

a = -30/t

plugin values into the first equation:

y = v₁t + 1/2 at²

49 = 30t + 1/2 (-30/t)t²

49 = 30t -15t

49 = 15 t

t = 49/15

t = 3.27 seconds

2.

y = v₁t + 1/2 at²

a = -30/3.27

a = 9.2

y = 30(3.27) + 1/2(9.2) 3.27²

y = 147.3 m