Question

If a rocket of mass 117 kg is launched with a velocity of 800 m/s, how high does it rise?

Answer 1

`h = 3.26xx10^4` `"m"`

Explanation:

We're asked to find the maximum height of a projectile (rocket) given its initial speed.

In idealized projectile motion, the only change in motion it expletives is the downward acceleration due to gravity, independent of its mass.

With that being said, we can use the equation

`(v_y)^2 = (v_(0y))^2 - 2gh`

where

• `v_y` is the velocity at a certain height `h`

• `v_(0y)` is the initial velocity, given as `800` `"m/s"`

• `g` is the acceleration due to gravity near earth's surface,`9.81` `"m/s"`

• `h` is the height, in `"m"`

At its maximum height, the instantaneous velocity `v_y` is `0`, so we have

`0 = (800color(white)(l)"m/s")^2 - 2(9.81color(white)(l)"m/s"^2)h`

`h = ((800color(white)(l)"m/s")^2)/(2(9.81color(white)(l)"m/s"^2)) = color(red)(3.26xx10^4` `color(red)("m"`

Answer 2

• First and foremost, calculate kinectic energy (I will just name it KE) of the rocket

`KE=1/2m(v^2)`, where,

m=mass of rocket (kg)

v=velocity of rocket (m/s)

Therefore,

`KE=1/2(117)(800^2)`

`KE=3,744,0000=3.744*10^7`(standard form so that we are less prone to errors occurring)

• Then, assuming that KE and GPE (Gravitational Potential Energy) are equal:

`GPE=mgh`

g= gravitational strength (for Earth 10N/kg)

h=height(m)

Therefore,

`3.744*10^7=(117)*(10)*h`

`3.744*10^7=1170h`

`1170h=3.744*10^7`

`h=(3.744*10^7)/(1170)`

`rarr h=32000m`