Question

Use the worked example above to help you solve this problem. A rocket moves straight upward, starting from rest with an acceleration of +29.5 m/s². It runs out of fuel at the end of 4.17 s and continues to coast upward, reaching a maximum height before falling back to Earth.

(a) Find the rocket's velocity and position at the end of 4.17 s.

vb =	m/s
yb =	m

(b) Find the maximum height the rocket reaches.
m

(c) Find the velocity the instant before the rocket crashes on the ground.
m/s

Answer 1

Ca 4,20 kx* L Ź x 29.5*(4-1792 motion from A to B duitial velocity (at A) a=0 By UB=123m/s time t = 4017 see acceleration of 29.5 m/s2 Y = At that? - 2 =o+ cat? ad u os surface = 256.49 m final velocity Up (at B7 = UA + at >> UB= 0 + at = 29.5x4.17=123 m/s motion from B to a Initial velocity Up=123 m/s final velocity Uc=0 acceleration a=-9--9-8 m/s2 ( g is acting downward) U² = u + 2aY₂ 7 0 = Us -zgyz >> Yz a ver 2 123)* = 771-89 m Total height achieved h=Y, +Y₂ (=CA) = 256-49+771-89 = 1028.38 m Let mass of rocket late) be m and at the point Al ground surface] its velocity be o » P.E at c= koE at A mgh = 1 m² = e = √2gh = /2x9.8X1028.38 = 141-97 m/s