SOLUTION: 1. The height of an object launched vertically above the surface of a planet is given by the function h=-1/2gt^2+vt+h, where g is the gravtiational acceleration on that planet, v i

Question 1167000: 1. The height of an object launched vertically above the surface of a planet is given by the function h=-1/2gt^2+vt+h, where g is the gravtiational acceleration on that planet, v is the initial velocity and h is the initial height of the object.
The gravitational acceleration on Mars is 4m/s^2. An astronaut jumps from a 10 metre high rock into the air with velocity of 8m/s upwards. He reaches a max height then accelerates downwards and eventually lands safely on the ground beneath the rock.
a)What is the max height reached by the astronaut
b)When is that max height achieved?
c)What is the duration of his flight?
d)When is the astronaut at the same height as the rock on his way down?
e)For how long is the astronaut higher than 12 metres above the ground?
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

You have slightly mis-stated the height function for an object near a massive body that has been given a positive initial velocity beginning at an initial height.

You said

Which is ludicrous because how can h be equal to itself plus a bunch of other non-zero stuff?

The initial height must be a different variable, and it makes sense to identify the initial velocity as something other than just because the velocity is constantly changing throughout the domain of the function.

Here is the expression that represents height as a function of elapsed time:

Where is the local constant representing the gravitational acceleration, is the initial velocity, and is the initial height.

You should recognize this as a quadratic function with a negative lead coefficient meaning that the graph is a convex down parabola. The maximum height is the value of the function at the time the vertex is reached, and the time the vertex is reached is the additive inverse of the coefficient of the first degree term, namely divided by two times the lead coefficient, namely .

So you need to answer part b first, as explained above. Then answer part a by finding the value of the function at the time calculated for part b. The duration of the flight, part c, is the positive zero of the function when it is equal to zero because he will be flying until his height is zero. For the last part, set the function equal to 12 and solve for the two positive zeros. The duration of the flight where he is above 12 meters is the difference between those two zeros.

John

My calculator said it, I believe it, that settles it

I > Ø

Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.

If you want to learn the subject to that level to solve such problems on your own and to learn it from a good source,
you have this lucky, happy and fortunate opportunity.

In this site, I prepared the lessons that cover the entire subject
    - Problem on a projectile moving vertically up and down
    - Problem on an arrow shot vertically upward
    - Problem on a ball thrown vertically up from the top of a tower
    - Problem on a toy rocket launched vertically up from a tall platform

Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems once and for all.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Projectiles launched/thrown and moving vertically up and dawn".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

DO NOT MISS YOUR CHANCE (!)

It is really a UNIQUE CHANCE (!)

RELATED QUESTIONS
A 1� oz. bottle of roll-on deodorant has a height of 4 in. and a radius of � in. Find the (answered by solver91311)
An object is launched upward from the groumf an initial velocity of 200 ft/sec. The... (answered by Glaviolette)
"An object is launched vertically in the air from a 10-meter tall platform. The height... (answered by Alan3354)
The height h in feet of an object after t seconds is given by the function h = -16t^2 + (answered by ikleyn)
Hi. My teacher gave me this problem and i dont understand it: The general formula for... (answered by jim_thompson5910,Earlsdon)
I am not sure if this is in the right section but the question is: On Planet X, the... (answered by nerdybill)
The formula for the height, h metres, of an object propelled into the air is h = -1/2g t (answered by scott8148)
The height of fireworks during a Fourth of July show can be modeled by quadratic... (answered by lynnlo)
(1.) The height of a triangle is 4 feet less than twice the base. If the area is 48... (answered by nerdybill)