Question

An object is launched from ground level with an initial velocity of 120 meters per second. For how long is the object at or above 500 meters (rounded to the nearest second)?

Answer 1

The height of the object can be described by the equation
$y= \frac{1}{2} g t^2 +v_0t+h_0$
where y is the height in meters
g is the gravitional constant which is equal to -9.81m/s^2
t is time
v0 is the initial velocity
h0 is the initial height

v0 is given to be 120
h0 is zero because it is launched from the ground and the ground has a height of 0m

now, the question is asking for the time in which the object remains above 500m
so, if you substituted all these values into the given equation
$500= \frac{-9.81}{2}t^2 + 120 t +0$

then you can solve for two time values in which the object is exactly at 500m (as its going up and as its falling down)

to find the amount of time, you just take the difference between the two time (subtract one from the other)


any questions?

Answer 2

The CORRECT answer is 14. :)