Solve for Projectile Motion: Maximum Altitude, Time of Flight, and Range

[Kildars]

A rocket is launched at an angle of 57.0° above the horizontal with an initial speed of 96 m/s. It moves for 3.00 s along its initial line of motion with an acceleration of 30.0 m/s^2. At this time its engines fail and the rocket proceeds to move as a free body.

(a) Find the maximum altitude reached by the rocket.
m
(b) Find its total time of flight.
s
(c) Find its horizontal range.
m

I tried finding \Delta X using

\Delta X = Vo^2/g2sin(theta)

I got 120.8117 for X

then i plugged it in Delta X = Vo X sin 2a / g

 

[Kildars]

bump

my message is too short so i am typing this to make it longer.

 

[Bernie Hunt]

Kildars,

Go back and look at the formula's I gave you on the first problem with the cliff. They should be all you need.

Max altitude is when y velocity is zero. Total flight time is when y location is zero. That will give you the flight time, plug that time into the x location formula and you are done.

Bernie

 

[johnnyp]

I think it's a little trickier than that. Kilders, here's a hint: during the first 3 seconds, the motion is linear (what can you deduce from that?), and then the rocket proceeds as a projectile (the final velocity of the linear motion is the initial velocity of the projectile).

Hope this helps. Ask if you need further clarification.