Question

A rocket moves straight upward , starting from rest with an acceleration of 29.4 m/s2. it runs out of fuel at the end of 4.00 s and countinues to coast upward , reaching a maximum height before falling back to earth .

A) find the rocket's velocity and position at the end of 4.00 s .
B) Find the maximum height the rocket reaches.
C) find the velocity the instant before the rocket crashes on the ground

Answer 1

Answer:

117.6 m/s, 235.2 m

940.08073 m

135.81 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity = 9.81 m/s² = a

$v=u+at\\\Rightarrow v=0+29.4\times 4\\\Rightarrow v=117.6\ m/s$

The velocity at the end of 4 seconds is 117.6 m/s

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 29.4\times 4^2\\\Rightarrow s=235.2\ m$

Position at the end of 4 seconds is 235.2 m above the ground

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-117.6^2}{2\times -9.81}\\\Rightarrow s=704.88073\ m$

Maximum height of the rocket is 704.88073+235.2 = 940.08073 m

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 940.08073+0^2}\\\Rightarrow v=135.81\ m/s$

Velocity of the rocket as it crashes is 135.81 m/s