Question

A rocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s but does not stop. m/s2 how high does it rise above the ground?

Answer 1

consider the motion of rocket until it runs out of fuel

v₀ = initial velocity = 0 m/s

v = final velocity when it runs out of fuel = ?

t = time after which fuel is finished = 3.98 sec

a = acceleration = 29.4 m/s²

Y₀ = height gained when the fuel is finished = ?

using the kinematics equation

v = v₀ + a t

v = 0 + (29.4) (3.98)

v = 117.01 m/s

using the equation

v² = v²₀ + 2 a Y₀

(117.01)² = 0² + 2 (29.4) Y₀

Y₀ = 232.85 m

consider the motion of rocket after fuel is finished till it reach the maximum height.

Y₀ = initial position = 232.85 m

Y = final position at maximum height

v₀ = initial velocity just after the fuel is finished = 117.01 m/s

v = final velocity after it reach the maximum height = 0 m/s

a = acceleration due to gravity = - 9.8 m/s²

using the kinematics equation

v² = v²₀ + 2 a (Y - Y₀)

inserting the values

0² = (117.01)² + 2 (- 9.8) (Y - 232.85)

Y = 931.4 m

Answer 2

U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s

Let h₁ =  the height at time t, for t ≤ 3.98 s
Let h₂ =  the height at time t > 3.98 s

Motion for  t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s

Motion for t  > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m

The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m

Answer: 931.4 m (nearest tenth)