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:
\ =\ -\frac{1}{2}gt^2\ +\ v_ot\ +\ h_o)
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):
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