Find space in the air also speed of rocket?

Question

When the rocket is launched upward into the air, we use the formulas
D=RT-5t^2
D = distance(in meters) above its start point,
T= time since it was launched,
R=initial speed of rocket in meter per second.
If rocket's launched upward from the ground with an intial speed of 45 meters per sec, what's the first time the rocket reaches a height of 100 meters? Round your answer o nearest tenth of second.

Answer 1

You have a T and a t...doesn't seem like much but in physics T can be period while t is time...this may mean something if you are considering curvature (orbital mechanics) or just a standard physics problem...

I am assuming this is an algebraic problem and if it is and T and t are the same...plug in 100 for D since you are asked to find it at that height and plug in 45[m/s] for R. Then you will have a quadratic equation from algebra...the smaller answer will be the first time it reaches 100[m]. I don't have a calculator so I will leave the calculating to you but it should look like this with factoring:

D=RT-5T² -> 100=45T-5T²

Now move all terms to one side and let the other side of the equation equal 0..it is usually best to have your leading term (the highest order variable) be positive, aka 5T², now you have:

5T²-45T+100=0

All of these terms are divisible by 5 so let's simplify it and make it look prettier by dividing the entire equation by 5 (remember 0/5=0 for the right hand side of the equation.)

5T²-45T+100=0 -> T²-9T+20=0

The formula is in the form ax²+bx+c, use the ac method where you multiply a (1) to c (20) and then determine all factors of that to see which add up to 9. a,b and c are the coefficients which are a=1, b=9 and c=20 for our equation. If confused why a=1, remember that we don't normally write a coefficient of 1. Coefficient is a fancy math way of saying the number in front of the variable...and the variable is a fancy math way of saying the unknown value we are looking for...which is T in this problem. I don't know where you stand so I wanted to clarify before moving forward with all of these crazy terms...ok, let's continue to factor:

T²-9T+20=0
1*20=20 ->
20=20*1..20 and 1 cannot add or subtract to be -9, let's try another
20=10*2..10 and 2 cannot add or subtract to be -9..so let's continue
20=5*4..5 and 4 can add up to be -9 so long as they are both negative...so let's use these!

T²-9T+20=0 -> (T-5)(T-4)=0, thus: T=5 or T=4, since 4 occurs first it is out answer! 

Turns out a calculator wasn't needed for this problem

 

 

Side note (if you are curious), the reason it reaches 100[m] twice is because the old saying...what goes up must come down. Since 4 and 5 seconds are so close we can assume it is close to it's maximum height at 100[m] and is beginning is descent at 5 seconds...assuming simple physics conditions...just some extra info if you were curious =)

I hope this helped!

v/r

Daniel

p.s. please excuse typos... :\

Answer 2

Hello again, T.

STAIR-STEP: 2 secs will not get the Rt high enough (only 90m - 5t*t) ... try 3 secs for 135 -45 ... need 10m more ... try 4 secs for 180 -80 for 100m exactly ==> 4.0 secs  Best :)

Answer 3

You are using a simplification of a kinematic equation. (A list can be found here: http://www.physicsclassroom.com/class/1dkin/u1l6a.cfm)

 

100 = 45t - 5t²

-5t² + 45t -100 = 0

5t² - 45t + 100 = 0

5(t² - 9t + 20) = 0

t² - 9t + 20 = 0

(t - 5)(t - 4) = 0

t = 4 sec , 5 sec