# A little problem help

1. Oct 22, 2006

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Before you get all worked up, im just gonna let you know that this isnt homework. Here is my problem : A projectile is fired at a 20 degree angle with a velocity of 15 m/s. What is the total time of flight? What is the distance of its hight? How far did it travel in the horizontal direction?

Ok, no to solve this i used Vf^2=Vi^2-2ad, (this is on the trip up) where Vf = 0, Vi = 15, and a= 9.8, to give me the vertical distance of 11.48m. To find the horizontal distance, i used 15cos20 as the intial velocity and plugged it into the same equation to get a hor. distance of 10.13m. Then, to find total time i used a= (Vf-Vi)/t, where on the trip up a=9.8, Vf =0, and Vi = 15 to get a time of 1.53 sec, on the way down i used a=9.8, Vf=15, and Vi = 0 in the same equation to get the same exact time, which combine gave me a total of 3.06s.....so all of my answers are as follows:
Dv=11.48m, Dh=10.13m, Ttotal=3.06s
If anyone could confirm or correct my process id be greatly appreciated. Also im new so if this is indeed the wrong board i would gladly be redirected Thanks in advance

Last edited: Oct 22, 2006
2. Oct 22, 2006

Use the 'standard' equations for projectile motion:

$$v_{x}(t) = v_{0}\cos(\alpha)$$
$$v_{y}(t) = v_{0}\sin(\alpha) - gt$$
$$x(t) = v_{0}\cos(\alpha)t$$
$$y(t) = v_{0}\sin(\alpha)t - \frac{1}{2}gt^2$$.

3. Oct 22, 2006

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Wow, my teachers hasnt given us those equations so im really not sure how to use them

4. Oct 22, 2006

How to use them? Simply read them.

(Hint: use the condition y(t) = 0 to get the time of the flight t. Then you can find all you need.)

5. Oct 22, 2006

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Wow!!!! Excuse my terrible english in the previous statement......"my teachers hasnt...", you must think im retarded

6. Oct 23, 2006