Question

A catapult launches a rocket at an angle of 49.6° above the horizontal with an initial speed of 112 m/s. The rocket engine immediately starts a burn, and for 3.58 s the rocket moves along its initial line of motion with an acceleration of 32.5 m/s². Then its engine fails, and the rocket proceeds to move in free-fall.

(a) Find the maximum altitude reached by the rocket.
m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m

Answer 1

= angle of launch = 49.6

V_i = initial speed of rocket = 112 m/s

a = acceleration = 32.5 m/s²

t = time = 3.58 sec

Distance covered by the rocket is given as

y = V_i t + (0.5) at²

y = 112 (3.58) + (0.5) (32.5) (3.58)² = 609.23 m

height gained before engine fails = H₁ = y Sin49.6 = 609.23 Sin49.6 = 464 m

horizontal distance travelled before engine fails of launch = R₁ = y Cos49.6 = 609.23 Cos49.6 = 395 m

velocity just before engine fails is given as

V_f = V_i + at = 112 + (32.5) (3.58) = 228.4 m/s

initial velocity just after engine fails along y-direction = V_oy = V_f Sin49.6 = 228.4 Sin49.6 = 174 m/s

initial velocity just after engine fails along x-direction = V_ox = V_f Cos49.6 = 228.4 Cos49.6 = 148 m/s

motion after engine fails ::::

height gained by rocket after engine fails can be given as

H₂ = V_f² Sin² / 2g = (228.4)² Sin² (49.6) / (2 x 9.8) = 1543.55 m

a)

Maximum altitude reached by rocket = H₁ + H₂ = 464 + 1543.55 = 2007.6 m

b)

Consider motion of rocket along Y-direction after engine fails

V_oy = 174 m/s

a_y = -9.8 m/s2

d = displacement = -H₁ = - 464 m

t' = time taken after engine fails

Using the equation

d = V_oy t' + (0.5) a t'²

-464 = 174 t' + (0.5) (-9.8) t'²

t' = 38 sec

Total time of flight = t + t' = 3.58 + 38 = 41.58 sec

c)

Horizontal distance covered after the engine fails = R₂ = V_ox t' = 148 x 38 = 5624 m

Total Horizontal distance travelled = R₁ + R₂ = 395 + 5624 = 6019 m