A rocket is launched at an angle of 52.0° above the horizontal with an initialsp

Question

A rocket is launched at an angle of 52.0° above the horizontal with an initialspeed of 96 m/s. The rocket moves for3.00 s along its initial line of motion with an acceleration of32.0 m/s2. At this time,its engines fail and the rocket proceeds to move as a projectile. (a) Find the maximum altitude reached by therocket.
(m)
(b) Find its total time of flight.
(s)
(c) Find its horizontal range.
(m) (a) Find the maximum altitude reached by therocket.
(m)
(b) Find its total time of flight.
(s)
(c) Find its horizontal range.
(m)

Explanation / Answer

Initial speed u = 96 m / s time t = 3 s acceleration a = 32.0m/s2 velocity after 3 s is v = u + at from thiswefind   v value (a). maximum altitude H = v 2 sin2 / 2g where = 52 o (b). total time of flight T = 2 v sin / g (c). Horizonatl range R = v 2 sin 2 /g substitue values we get answer velocity after 3 s is v = u + at from thiswefind   v value (a). maximum altitude H = v 2 sin2 / 2g where = 52 o (b). total time of flight T = 2 v sin / g (c). Horizonatl range R = v 2 sin 2 /g substitue values we get answer

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