Question

a rocket is launched straight up with a speed of 12 meters per second near the surface of earth. [neglect air friction.] [show all work, including the equation and substitution with units.] a) calculate the maximum height reached by the rocket; b) calculate the time the rocket needs to return back to the launch point.

Answer 1

a) 7.34 meters

b) 2.446 seconds

Step-by-step explanation:

Part a)

First take note.

Initial velocity = 12

Initial horizontal velocity = 0

Initial vertical velocity = 12

If something is launched straight up in the air, it is launched at angle of 90 degrees.

`h_(max) =(v_(0y) ^(2) )/(2g)`

This equation defines the maximum height of a projectile above its launch position.

`h_(max)` = maximum height

`v_(0y) ^(2)` = initial velocity in the y direction

`g` = gravity

We are given

`v_(0y) ^(2)=12`

`g=9.81`

Lets solve for
`h_(max)`.

`h_(max) =(12 ^(2) )/(2*9.81)`

`h_(max) =(144 )/(2*9.81)`

`h_(max) =(144 )/(19.62)`

`h_(max) = 7.34` meters

Part b)

`T_(tot) =(2(v_(0)*sinx) )/(g)`

Note x = angle

This is the time of flight for a projectile both launched and impacting on a flat horizontal surface.

We are given.

`v_(0)=12`

`x=90`

`g=9.81`

Lets solve for
`T_(tot)`.

`T_(tot) =(2(12*sin90) )/(9.81)`

`T_(tot) =(2(12*1) )/(9.81)`

`T_(tot) =(2(12) )/(9.81)`

`T_(tot) =(24 )/(9.81)`

`T_(tot) =2.446` seconds