# A rocket is launched at an angle of 53 degrees abo...

A rocket is launched at an angle of 53 degrees above thehorizontal with an initial speed of 100m/s. The rocket moves for3.00s along its initial line of motion with an acceleration of 30.0m/s/s. At this time, its engines fail and the rocket proceeds tomove as a projectile. Find (a) the maximum altitude reached by therocket, (b) its total time of flight, and (c) its horizontalrange.

1. okay
lets find the maximum altitude of the rocket
lets not worry about the components until the rocket has doneaccelerating
the angle will remain constant duringacceleration.
the velocity the second the rocket stops accelerating is
Vf = Vi + a t
Vf = 100 + 30 (3)
Vf = 190 m/s still acting at 53 degrees.
how far has it moved in each direction
d = V1t + 1/2 at2
d = 100(3) + 15(9)
d = 435m
lets look at the drawing so we can do some basic trig to find its height and horizontaldistance
Sin = opp / hyp   sin 53 = height /435   height = 357.4 m

Cos = adj / hyp     cos53 = distance /435    distance = 261.7 m

alright!   now we have the height and distance when itstops accelerating.  Now its a simple projectile motionquestion

Going back we found it had a velocity of 190 m/s

lets split that into x and y component

Velocity y = sin53 * 190   = 151.7m/s
Velocity x = cos53 * 190 = 114.3 m/s

a) max altitude    just concerned with ycomponent
now our old Vf is our new Vi   when it reaches itsmaximum height the velocity final will be zero since gravity isaccelerating it down

Vf = Vi + at
0 = 151.7 - 9.8t      t = -151.7 /-9.8   t = 15.4s

again

d = V1t + 1/2 at2
d = 151.7(15.4) - 4.9 (15.4)^2
d = 2336m - 1162m = 1174m

But that was after the engines remember before the enginesblew it had a height of 357.4 m

so total height = 1174 + 357.4 = 1531.6 m

b) so right now we know the total time to its max height
but we still need to find the time it takes to free fall toearth

d = V1t + 1/2 at2
V1 = 0 now

d = 4.9 t^2
t^2 = 1531.6 / 4.9 = 312.6
t = 17.6 s

so lets add up the total times

3s till engine fall , 15.4s till it reaches max alt and 17.6seconds down

Total time = 36 s

c) horinzontal range

once the engines blew the x velocity was constant.

recall

Velocity x = cos53 * 190 = 114.3 m/s

so d = vt
d = 114.3 (33)   the first 3 seconds werent constant

d = 3771.9

but we must add on the original distance when engine blew

distance = 261.7 m

so total distance = 3771.9 + 261.7 = 4033.6 m

there you go

hope it helped