Two equations are needed.

v = u + at

(u-initial velocity, a-acceleration, t-time, v-final velocity)

s = ut + 0.5at^2 (s= distance)

Velocity at end of 8 secs = 0 + 49 x 8 = 392 m/sec

Distance travelled in 8 secs = 0 x 8 + 0.5 x 49 x 64

= 1568 m

From 392 m/sec the rocket decelerates to zero.

0 = 392 – 4.9t; or t = 40 secs.

From 392 m/sec the rocket casts to zero in 40 secs.

Distance = 392 x 40 – 0.5 x 4.9 x 1600

(minus because it is decelerating)

= 7840 m

Total height rockety reaches = 1568 + 7840

= 9408 m

Height in the vertical direction is given by

y = y_o + v_ot + 1/2at^2.

This will be a two part problem The first part is to solve for the height for the acceleration period. The second part is to find the height after the acceleration ends.

Part 1)

Using the above equation you find the first height.

y = 0 + 0 + 1/2 at^2

y_o = 0 because the height is 0 at start.

v_o = 0 because you are starting from rest.

So now you have:

y = 1/2 * 49 * 8^2 = 1568 m

Now we have to find the height the rocket goes after the acceleration stops. It will have a velocity because the acceleration has boosted its speed from rest.

After 8 seconds the velocity should be

v = v_o + at

v = 0 + 49 * 8

v = 392 m/s

Now that we have the speed of the rocket after 8 seconds we can figure out what the height is after the rocket stops accelerating. We will use the original equation again but set the starting point at the moment the acceleration stops.

The problem is how do we know when it stops going up? We know the rocket will stop going up at the point the velocity reaches 0. When the velocity reaches 0 then the rocket will reverse direction and start dropping back down to earth. So before we can even find the height we need find out the time it takes for the rocket to reach that point.

v = v_o + at

v = 0 because of the reason above. v_o is the velocity at the end of the acceleration. a is given as 9.8 m/s^2. Remember that a is negative because gravity pulls downward.

0 = 392 -9.8t

solve for t.

t = 392/9.8 = 40 s.

So now we know the rocket kept going up for 40 seconds. We can know figure out the height.

y = y_o + v_ot + 1/2 at^2

y_o is now the height from the end of part 1. v_o is the velocity after part 1. Since the rocket is not accelerating upward anymore, we will use the acceleration due to gravity for a. Plug in 40 seconds for the time.

y = 1568 + 392(40) + 1/2 (9.8) (40)^2

y = 1568 + 15680 + 7840

y = 25088 m

I hope that’s right!

Hello Linc, I am going to use the suvat versions of the equations of motion. You must do this problem in two parts. First you consider only from the launch to the point that the engine cuts off. You must find the speed right when the engine stops. Then you use this speed to find the additional distance traveled without the engine (which is just like any object thrown upward) In the first part Distance when the engine stops = s = 150 m Initial velocity = u = 50 m/s Acceleration = a = 2.0 m/s^2 m/s^2 We have three variables, that is all we need to solve for the other two, v or t. Here we need v, the velocity when the engine stops. so we need a suvat equation with u, s, a, and v, but not t. v^2 = u^2 + 2as v = sqrt((50 m/s)^2 + 2 * 2 m/s^2 * 150 m) v = 55.7 m/s Now our next part considers only the flight AFTER the engine stops. For this part the initial velocity (u) is the final velocity (v) from the last part. u = 55.7 m/s v = Final velocity (at the peak of flight the vertical velocity is zero) = 0 m/s a = – 9.81 m/s^2 There you go, three variables, and you need to find s, so again we need the equation of motion WITHOUT time, but with s, u, v and a v^2 = u^2 + 2as solve for s s = (v^2 – u^2) / 2a but v is zero so get rid of it. s = – u^2 / 2a s = – (55.7 m/s)^2 / (2 * -9.81 m/s^2) s = 158.1 m But that is only the distance after the engine shut off, add the 150 m from before and the final answer is Height = 308.1 m Good Luck Linc Sorry, I didn’t notice the part about the Time. There are three times here, lets call them t1, t2 and t3 t1 is the time till the engine stops t2 is the time from then til it reaches the peak of flight. t3 is the time to fall back to the ground From our data from the first part. a = 2.0 m/s^2 u = 50 m/s v = 55.7 m/s s = 150 m (Forget about that stupid equation with the t squared, we have enough data to choose ANY suvat equation, why not choose the easiest.) v = u + at t1 = (v – u) / a t1 = (55.7 m/s – 50 m/s) / 2.0 m/s^2 t1 = 2.85 sec From part two above. s = 158.1 m u = 55.7 m/s v = 0 m/s a = – 9.81 m/s^2 Again you have FOUR variables now, you can chose any suvat equation, I would pick one with v, because it is zero and cancels out v = u + at t2 = (v – u) / a t2 = (0 m/s – 55.7 m/s) / – 9.81 m/s^2 t2 = 5.68 sec And for t3, consider the part from the peak of flight back to the ground. Initial velocity = u = zero a = – 9.91 m/s^2 s = 308.1 m Ok, with only three variable we need an equation with u, a, s and t s = ut + (1/2) at^2 This equation is not too bad to solve for t if the initial velocity is zero. s = (1/2) at^2 solve for t t3 = sqrt(2s/a) t3 = sqrt(2 * – 308.1 m / – 9.81 m/s^2) t3 = 7.93 sec Now just add t1 and t2 and t3 Total time in air = 16.46 sec Check may math of course, and I sympathize with having a bad teacher, that is the worst. Good Luck

Rocket Physics Problem