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.

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## Answers ( 1 )

_{f}= V_{i}+ a t_{1}t + 1/2 at^{2}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

V

_{f}= V_{i}+ at0 = 151.7 - 9.8t t = -151.7 /-9.8 t = 15.4s

again

d = V

_{1}t + 1/2 at^{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 = V

_{1}t + 1/2 at^{2}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

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