Find maximum height and range of a rocket that is launched at 70 degrees.

Question

During 6.5 seconds, a rocket rises at a rectilinear trajectory oriented to 70 degrees, with a constant acceleration of 8 m/s^2. It is in free fall. Find a) the maximum height and b) the range.

Much appreciated!

Answer 1

Solution:

a) The max height of the rocket before it starts falling can be found from y = -1/2 gt²+ V_o sin ∝ t + y_o

the time for the rise is t = 6.5 sec, and the angle of the launch is ∝ =70º

the acceleration is constant a = 8 m/s² and the derivative of the initial velocity is dV_o/dt = a

therefore dV_o = a dt taking integral for both sides you get V₀ = a t , and V_o = 8 m/s²( 6.5 s) = 52 m/s

substituting in the equation above and knowing that y_o = 0 when the rocket was still on the ground

y = - 1/2 ( 9.8 m/s²) ( 6.5 s) + (52 m/s) sin 70º ( 6.5 s) = -31.85 m + 317.62 m = 286 m

b) the range is the value x = V_o cos ∝ t = 52 m/s cos 70º ( 6.5 s) = 115.6 m