A signal rocket is launched vertically at a velocity of 80 m/s, working out several figures?

Question

A signal rocket is launched vertically at a velocity of 80 m/s
Calculate:
a) The time taken to reach its maximum height
b) The maximum height reached
c) The velocity at a height of 27m
d) Its height after 5 seconds

Answer 1

a) when the rocket reaches its maximum height, its final veloicity will be 0m/s
Vf = at + Vi
0 = (-9.8)t + 80
-80 = -9.8t
t = 8.163s

b) again, its final velocity will be 0m/s at maximum height
Vf^2 = 2ad + Vi^2
0^2 = 2(-9.8)d + 80^2
-6400 = -19.6d
d = 326.53m

c) Vf^2 = 2ad + Vi^2
Vf^2 = 2(-9.8)(27) + 80^2
Vf^2 = -529.2+6400
Vf^2 = 5870.8
Vf = 76.62m/s

d) X = 1/2at^2 + vt + xi
X = 1/2(-9.8)(5)^2 + 80(5) + 0
X = -122.5 + 400
X = 277.5m

Answer 2

First, some assumptions

one is that the only force acting on the rocket is gravity, that whatever accelerated the rocket to 80 m/s did so instantaneously
Also, no air resistance
Finally, purely vertical motion.

Under these assumptions, a projectile follows the equations:
Speed
v(t)=v0+a*t

Position
s(t)=s0+v0*t+.5*a*t^2

in this case, s0=0, v0=80, and a=-9.81

a). The max height is achieved when v(t)=0
0=80-9.81*t
t=80/9.81
t= 8.2 s

b) knowing t at apogee
s(8.2)=80*8.2-.5*9.81*8.2^2
=326 m

c) First, evaluate s(t)=27
27=80*t-.5*9.81*t^2

Note that this is a quadratic sine the projectile will pass through 27 m on the way up and have positive v, and on the way down with a negative v.

the roots are the times
15.97 and 0.345 seconds

The velocity at the times should have equal magnitude, opposite sign
v(15.97)=80-9.81*15.97
=-77 m/s
v(0.345)=80-9.81*0.345
=77 m/s

yup

d)
this is just s(5)
s(5)=80*5-.5*9.81*25
=277 m

j