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)?

A) 11 sec
B) 12 sec
C) 13 sec
D) 14 sec

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)?

A) 11 sec
B) 12 sec
C) 13 sec
D) 14 sec

anonymous · Answer

Please help Rick.

anonymous · Answer

lol

I think you need to use the kinematic equations for free-fall for this problem.  Is it launched perfectly upward with the initial velocity, and then only responding to acceleration due to gravity at free-fall?

anonymous · Answer

I dont understand

shamim · Answer

can u tell me the velocity at 500m height

anonymous · Answer

Is there any more information given, or is that all?

The kinematic equations for free-fall are:

$$d = v_i(t) * \frac{1}{2}*(-9.8 m/s)t^2$$

and so on.  Have you seen these?

anonymous · Answer

Sorry, that should be a + 1/2(-9.8).... not *

anonymous · Answer

They only other thing they gave me in the question was this hint: y = -4.9t^2 + initial velocity * t + initial height

anonymous · Answer

So y there is height (distance).

anonymous · Answer

First, plug in 500 for y and 120 for initial velocity:

$$500 = -4.9t^2 + 120(t)$$

anonymous · Answer

(initial height is zero)

anonymous · Answer

Now, solve for t, which will be the time in seconds that it took to reach 500m.

anonymous · Answer

$$4.9t^2 -120t + 500 = 0$$

anonymous · Answer

I guess you could use the quadratic equation, but I'm going to cheat and plug it into a calculator.

anonymous · Answer

Sorry the answers i'm getting don't match the choices

anonymous · Answer

Once you know t from that, I think you could use this equation:
$$V_f = V_i + -9.8t$$
To find the time at the maximum height of the projectile, when Vf = 0.
Then I think you would subtract your time at height = 500m from the calculated time here.

anonymous · Answer

$$0 = 120 + -9.8t \rightarrow 120 = 9.8t \rightarrow \frac{120}{9.8} = t$$

anonymous · Answer

so the answer is B

anonymous · Answer

So max height is t = 12.24 seconds.  Time at 500m = 5.32 seconds.    12.24 - 5.32 = 6.92 seconds above 500m, but that is only at the upward point.  It will will be above 500m from max meight to 500m when it starts falling.

anonymous · Answer

Is it B?

anonymous · Answer

I would have guessed 14seconds above 500m based on the calcs we just did, but sorry if I'm wrong..

anonymous · Answer

It's D thanks for your help.

anonymous · Answer

Woohoo. :)

anonymous · Answer

If you have more questions, here's what I used to research: http://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations-and-Free-Fall