Question

A rocket moves straight upward, starting from rest with an acceleration of +29.4 m/sec2 . It runs out of fuel after 4 seconds and continue to rise, reaching a maximum height before falling back to Earth. a) Find the rocket's velocity and height at the moment fuel ends. b) Find the maximum this rocket can reach. c) Find the velocity the instant before the rocket crashes on the ground.

Answer 1

Answer:

(a). The rocket's velocity is 117.6 m/s.

(b). The rocket can reach at maximum height is 940.8 m.

(c). The velocity the instant before the rocket crashes on the ground is 135.7 m/s.

Explanation:

Given that,

Acceleration = 29.4 m/s²

Time = 4 sec

(a). We need to calculate the rocket's velocity

Using equation of motion

$v=u+at$

Put the value into the formula

$v=0+29.4\times4$

$v=117.6\ m/s$

We need to calculate the maximum height at the moment fuel ends

For the value of x₁

Using equation of motion

$x_{1}=ut+\dfrac{1}{2}at^2$

Put the value into the formula

$x_{1}=0+\dfrac{1}{2}\times29.4\times4^2$

$x_{1}=235.2\ m$

We need to calculate the value of x₂

Using equation of motion

$v^2=u^2-2gx_{2}$

Put the value into the formula

$0=u^2-2gx_{2}$

$x_{2}=\dfrac{u^2}{2g}$

Put the value in to the formula

$x_{2}=\dfrac{117.6^2}{2\times9.8}$

$x_{2}=705.6\ m$

(b). We need to calculate the maximum this rocket can reach

Using formula for height

$H=x_{1}+x_{2}$

Put the value into the formula

$H=235.2+705.6$

$H=940.8\ m$

(c). We need to calculate the velocity the instant before the rocket crashes on the ground

Using equation of motion

$v^2=u^2+2gh$

Put the value into the formula

$v=\sqrt{2\times9.8\times940.8}$

$v=135.7\ m/s$

Hence, (a). The rocket's velocity is 117.6 m/s.

(b). The rocket can reach at maximum height is 940.8 m.

(c). The velocity the instant before the rocket crashes on the ground is 135.7 m/s.