A projectile travels above level ground. The initial vertical component of its velocity is 0.90 m/s and the initial horizontal component of its velocity is 0.60 m/s. Calculate the time the projectile remains in the air and the horizontal distance it travels.

Question

anonymous · Answer

Since we know the velocity of the projectile and its "x-component" velocity we can find the "y-component" easily by:

$$v _{y} = \sqrt{v - v _{x}}$$

after you get your "y-component velocity" plug it into this equation to get your time in flight. We set "y= 0" since that is when it hits the ground right?

$$y _{f} = y _{0} + v _{y}t + \frac{ 1 }{ 2 }at ^{2}$$

where "a" is our gravitational constant (-9.81 m/s^2). "t" is our time and y_0 is our beginning initial height. Solve for "t".

After you get your time in flight, it should allow you to solve the distance that it flew by using this equation:
$$x _{f} = v _{x}t$$Remeber that that the "x-component" and the "y-component" velocity and acceleration are independent of each other. Meaning they won't affect each other. So in the last equation remember to use the "x-component" of the velocity and plug in your "t" that you found from your first question. Since after all when it's in flight it's still moving right? meaning the distance is increasing until it hits the floor then it stops moving.

anonymous · Answer

This is how I solved it but I don't think it's right!
|dw:1382384599609:dw|

anonymous · Answer

Well seems like I read the question incorrectly since the horizontal and vertical velocity is already given. You can basically skip the first part and begin at the second equation.

anonymous · Answer

You have your image set up right. Have you ever seen those equations that I just gave you? Do you know how to work them?

anonymous · Answer

@OrionsBelt yeah I think i chose the wrong equation! I'll try that and let you know how it goes

anonymous · Answer

Okay. Well I think @DemolisionWolf is giving you another route. But try my way first if that doesn't work lets see what he has.

anonymous · Answer

Remember to skip the first part of my response as that is incorrect. You already have the velocity components.

anonymous · Answer

yf=y0+vyt+12at2
@OrionsBelt i'm confused, is yf the final velocity of y? and what's y0? I've used this equation before, i don't know if it's similar to the one you gave me?  d= v*t+1/2 at^2

anonymous · Answer

Imagine an image as so, where "y" is the height of the projectile in and "x" is the distance it travels:|dw:1382385444376:dw|