A rocket is launched at an angle of 53 degrees above the horizontal with an intial speed of 100 m/s. It moves along its initial line of motion with an accel. of 30 m/s^2 for 3 s. At this time its engines fail and the rocket proceeds to move as a free body. Find the maximum altitude reached by the rocket.
Just the four basic kinematics equations. . ?
The Attempt at a Solution
where is the rocket when its engines fail?
d= vit + (1/2)at^2
= 100 sin 53 (3) + (1/2)(30)(3^2)
= 374.59 m
what is the rocket's speed after the engines fail?
vf= vi + at
=100 + 30 (3)
So I feel like from here I should just be able to plug into vf^2= vi^2 + 2ad and find "d," using vi= 190 m/s and vf= 0. . . .But that's not the case. Why can't I do that? And where can I go from here? I have a feeling I need to find some way to express time to reach the maximum altitude, but I have no idea how I can express that.
~Thank you in advance!!