Fun physics problem, or so I thought..please help me :(?

You need to split it into 2 parts. The part where the engine accelerates it up and the part where it is just accelerated by gravity

For the first part, use the formula y = y0 + (v0)*t + 1/2 * (a^2) * t
y0 = 0m ( your starting height is 0)
v0 = 60 m/s (your starting speed)
y = 170m (your final height for part 1)
a = 2.5 m/s^2 acceleration
This gives you everything but t. You can solve for t to get the time it takes to finish part 1.
Then you can use the formula v = v0 * t
Plugging in your starting speed and the time that you just found gives you the speed at the end of part 1.

Now you have what you need for part 2.
Use the final speed and height from part 1 as the start of part 2.

A model rocket is launched straight upward with an initial speed of 60.0 m/s. It accelerates with a constant upward acceleration of 2.50 m/s2 until its engines stop at an altitude of 170 m.

(a) What is the maximum height reached by the rocket?

(b) How long after lift-off does the rocket reach its maximum height?

(c) How long is the rocket in the air?



The rocket has an initial altitude, d0, of 0 meters;
An initial velocity, v0, of 60 m/s;
And an initial acceleration, a0 of 2.50 m/s2, which is constant.

These conditions are at t0.

First find tb, or time at burn out (170m).
The equation of motion you’ll need is db = d0+v0t + ½ a0tb2.
Then solve:
db = d0+v0t + ½ a0tb2
170m = 0m + 60 m/s tb + 1.25 tb2
use the quadratic formula or solve numerically with excel, you’ll find it attains an altitude of 170.4 m in 2.69 s.

Now, you need to find the maximum height. This will be when its velocity equal zero. You again have a constant acceleration, but this time it is negative, due to gravity. I will say that ag = 9.81 m/s2. You again need to calculate the time, in this case to perigee, tp. You’ll need the formula vb = v0 + atb, and also again for vp = vb +atp. Now, its just “plug and chug”.

vb = v0 + atb
vb = 60 m/s + 2.50 m/s2 (2.69s)
vb = 60 m/s + 6.725 m/s (canceling the seconds)
vb = 66.73 m/s

Now to calculate the time to perigee, where we know that velocity is zero
vp = vb + atp
0 = 66.73 m/s – 9.81 m/s2 tp.
9.81 m/s2 tp= 66.73 m/s
tp = 6.80 s


Now going back to the distance formula:
dp = db+vbtp + ½ abtp2, making sure you have all your times correct.
dp = 170 m + 66.73tp – 4.91tp2
dp = 170 m + 66.73tp – 4.91tp2
dp = 396.93 m

This of course assumes the drag on the model rocket is negligible. Maintaining that assumption and not using a parachute, we use a distance formula again:

where dc is distance at crash, or 0 and tc is time to crash. Remember vp is 0.
dc = dp+vptc + ½ agtc2
0 = dp+ 0 + ½ agtc2
0 = 396.93 m – 4.91m/s2 tc2
396.93 m = 4.91m/s2 tc2
80.92 = tc2
tc = 9.00s

Now, the total time the rocket was in the air is
tb + tp + tc = 2.69s + 6.80s + 9.00s = 18.49 s

Ah silly mortals... you do no longer choose his weight! Gravity has a persevering with ACCELERATION of 9.8 m/s2, no rely how fat you're. a) substitute in place = preliminary velocity * substitute in time + a million/2*acceleration*substitute in time^2 consequently: 200m = (0m/s)(substitute in time) + a million/2 * 9.8m/s2 * (substitute in time)^2 resolve this to get (substitute in time)^2 = 40.8 s^2 substitute in time = 6.39s b) very final velocity = preliminary velocity + acceleration * substitute in time = 0m/s + 9.8 m/s2 * 6.39s = sixty two.6 m/s remark to jbs: that is purely unfavorable in case you define down as unfavorable... the right answer is sixty two.6 m/s down.

First, the initial speed of a rocket is always zero.
Second, the rocket will use a parachute on the way down. We cannot calculate total time in the air without knowing when it deploys.

I'll ignore that and take a shot at the question as asked.

While accelerating up:
s = v1t + 1/2at^2
s = position = 170
v1 = initial speed = 60.0
a = acceleration = 2.50
t = time
170 = 60t =1/2(2.50)t^2
solve for t
speed = v1 + at = 60.0 + 2.5t

After burnout while slowing due to gravity:
s = v1t = 1/2at^2
v1 = speed from above
a = -9.8(gravity)
t = v1 / 9.8( time for gravity to reduce speed to 0)
solve for s.
Add to 170 to find total height.

Falling to earth:
s = 1/2at^2

s = total height
a = 9.8
solve for t.
Add all times for total time in air

I have ignored 'terminal velocity' and air resistance.

How can it launch with an initial speed of greater than zero? Did it launch from underground? This is an impossible question otherwise.

Solve it the normal way, but bring this fact up to your teacher and you should get extra credit. If not you have a stupid teacher.

let me start with the (B) part,
now one of the equations for motion is
v=u+at
v=final velocity
u=initial velocity
a=acceleration
t=time
now any body thrown upwards comes to zero velocity at the end.
so,
0=60+2.5xt
-60=2.5t
-60/-2.5=t
24s=t

distance=5000m.

392.5m
7.11s
8.95s

do your own homework

170 m

2 s

2 s too