Question

A rocket is fired straight upward, starting from rest with an acceleration of 25.0 m/s^2. It runs out of fuel at the end of 4.00 s and continues to coast upward, reaching a maximum height before falling back to earth. (a) Find the rocket height and velocity when it runs out of fuel. (b) Find the maximum height the rocket reaches. (c) Find the velocity the instant before the rocket crashes into the ground. (d) Find the total elapsed time from launch to ground impact.

Answer 1

a) 200m, 100m/s

b) 710.20m

c) -117.98 m/s

d) 26.24 s

Step-by-step explanation:

To solve this we have to use the formulas corresponding to a uniformly accelerated motion problem:

`V=Vo+a*t` (1)

`X=Xo+Vo*t+(1)/(2)*a*t^2\\` (2)

`V^2=Vo^2+2*a*X` (3)

where:

Vo is initial velocity

Xo=intial position

V=final velocity

X=displacement

a)

`X=0+0*4+(1)/(2)*25*4^2`

the intial position is zero because is lauched from the ground and the intial velocitiy is zero because it started from rest.

`X=200m`

`V=0+25*4`

`V=100m/s`

b)

The intial velocity is 100m/s we know that because question (a) the acceleration is -9.8
`(m)/(s^2)` because it is going downward.

`0=100^2+2*(-9.8)*X\\X=510.20\\totalheight=200+510.20=710.20m`

c)

In order to find the velocity when it crashes, we can use the formula (3).

the initial velocity is 0 because in that moment is starting to fall.

`V^2=0^2+2*(-9.8)*(710.20)\\V=-117.98 m/s`

the minus sign means that the object is going down.

d)

We can find the total amount of time adding the first 4 second and the time it takes to going down.

to calculate the time we can use the formula (2) setting the reference at 200m:

`-200=0+100*t+(1)/(2)*(-9.8)*t^2`

solving this we have: time taken= 22.24 seconds

total time is:

total=22.24+4=26.24 seconds.