Question

A cart carrying a vertical missile launcher moves horizontally at a constant velocity of 30.0 m/s to the right. It launches a rocket vertically upward. The missile has an initial vertical velocity of 40.0 m/s relative to the cart.
(a) How high does the rocket go?
(b) How far does the cart travel while the rocket is in the air?
(c) Where does the rocket land relative to the cart?

Answer 1

Answer

given,

constant velocity of the cart (v)= 30 m/s

initial vertical velocity of missile (v₁)= 40 m/s

At maximum height of vertical velocity of the object is zero

a) maximum height of the rocket

    using equation of motion

     $v_2^2 = v_1^2 + 2 a y$

     $y = \dfrac{v_2^2-v_1^2}{2 g}$

     $y =\dfrac{0^2- 40^2}{- 2\times 9.8}$

          y = 81.63 m

b) calculation of time of flight

     $(y -y_0) = v_1 t - \dfrac{1}{2}gt^2$

     $0 =v_1 t - \dfrac{1}{2}gt^2$

     $t = \dfrac{2 \times v_1}{g}$

     $t = \dfrac{2 \times 40}{9.8}$

            t = 8.163 s

 distance travel by the cart = v x t

                     = 30 x 8.163

                     = 244.89 ≈ 245 m

c) rocket will land into the cart because there is no horizontal acceleration so, velocity remain same.