A model rocket Was built last weekend. We launched it in a nearby field. The rocket took off Straight up with an initial velocity 62.3 m/s. If it accelerates upward at 4.30 m/s2 until the engines stop 3.73 s later: (a) What is the maximum height the rocket reaches? (b) How long is the rocket in the air? (c) What is the velocity of the rocket as it hits the ground?

a)  y_total = 575.4 m , b)   t_total = 22.56 s , c)  v = 106.20 m

Explanation:

a) We can solve this exercise using the scientific relations

Let's find the height reached while the engines are running

         v₁ = v₀ + a t

         v₁ = 62.3 + 4.30 3.73

         v₁ = 78.339 m / s

in this time it rose

        y₁ = v₀ t₁ + ½ a t₁²

        y₁ = 62.3 3.73 + ½ 4.30 3.73²

         y₁ = 262.29 m

At this moment the engines are turned off and the rocket continues in a vertical launch, suppose that the acceleration of gravity is constant in this path

        v₂ = v₁ 2 - 2 g y₂

at maximum height the velocity is zero (v₂ = 0)

        y₂ = v₁ 2 / 2g

        y₂ = 78.339 2/2 9.8

        y₂ = 313.11 m

the total height is

       y_total = y₁ + y₂

       y_total = 262.29 + 313.11

       y_total = 575.4 m

b) the time the rocket is in the air is

       t_total = t₁ + t₂

where t₂ is the time after the engines have shut down

       y = y₁ + v₁ t₂ - ½ g t₂²

      y = 0

      0 = 262.29 + 78.339 t₂ - ½ 9.8 t₂²

      t₂² - 15.99 t₂ - 53.53 = 0

      t₂ = [15.99 ±√ (15.99² +4  53.53)] / 2

      t₂ = [15.99 ± 21.67] / 2

      t₂´ = 18.83 s

      t₂´´ = -2.84 s

since the time must be a positive quantity the correct result is t2 = 18.83 s, the total time in the air is

      t_total = 3.73 + 18.83

      t_total = 22.56 s

c) the speed when it hits the ground

We can perform this calculation starting with the maximum height y = 575.4 m where it has zero initial velocity (vo = 0)

      v² = v₀² + 2 g y

      v = √ 2gy

       v = √ (2 9.8 575.4)

       v = 106.20 m