Question

A model rocket is fired straight up from the top of a 55-m-tall building. The rocket has only enough fuel to burn for 3.0 s. But while the rocket engine is burning fuel, it produces an upward acceleration of 60 m/s2. After the fuel supply is exhausted, the rocket is in free fall and just misses the edge of the building as it falls back to the ground. Ignoring air resistance, calculate 1. Determine the height above the ground of the rocket when its fuel runs out. Express your answer with the appropriate units.

Answer 1

here,

the maximum height of building , h = 55 m

time taken , t = 3 s

the upward acceleration , a = 60 m/s^2

after the end of t1 = 3 s

the height above the building , h1 = 0 + 0.5 * a * t1^2

h1 = 0 + 0.5 * 60 * 3^2 m = 270 m

the final vertical velocity , v1 = 0 + a * t1

v1 = 0 + 60 * 3 m/s = 180 m/s

the height above the ground of the rocket when its fuel runs out , h = height of building + height traveled while accelerating + height traveled while in free fall

h = 55 + h1 + (v1^2 /2g) m

h = 55 + 270 m + (180^2 /(2*9.81)) m

h = 1976.4 m