A catapult launches a rocket at an angle of 52.29 above the horizontal with an i

Question

A catapult launches a rocket at an angle of 52.29 above the horizontal with an initial speed of 117 m/s. The rocket engine immediately starts a bun, and for 2.87 s the rocket moves an its initial line of motion with an acceleration of 35.7 m/s. Then its engine fails, and the rocket proceeds to move in free-fall (a) Find the maximum altitude reached by the rocket differs from the correct answer by more than 10%. Double check your calculations, m (b) Find its total time of flight Your response differs from the correct answer by more than 10%. Double check your calulations. ) Find its horizontal range.

Explanation / Answer

during engine burns:

r1 = (117 x 2.87) + 35.7 x 2.87^2 /2 = 482.8 m

v = 117 + (35.7 x 2.87) = 219.5 m/s

after that, vertical component = 219.5 sin52.2 = 173.4 m/s

at maximum, it will become zero.

0^2 - 173.4^2 = 2(-9.8)(h)

h = 1534.2 m

H_max = h + 482.8sin52.2 = 1915 m .....Ans

(b) after engine shut,

yi = 482.8 sin52.2 = 381.5 m

vertical dispalcement = 0 - 381.5 = - 381.5 m

v0y = 173.4 m/s

ay = - 9.8 m/s^2

y = v0y t + ay t^2 /2

- 381.5 = 173.4t - 4.9t^2

t = 37.5 sec

total time = 2.87 + t = 40.4 sec

(b) x = (219.5cos52.2)t + 482.8cos52.2

x = 5341 m

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