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How to find initial velocity in projectile motion problem..? 
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#1
Dec208, 05:58 PM

P: 5

I'm in a physics 1 honors course at my highschool and we have been working on projectile motion problems. I'm very confused on how to solve this problem though more because of the math than the physics I think.
The problem asks for the initial velocity of a spear that was thrown at a 35 degree angle with respect to the horizontal. The only other information given is that the spear was thrown a total horizontal distance of 201.24 meters. I first used the equation d_{y}=V_{iy}t+1/2a_{y}t^{2} and eventually came to t=V_{i}sin35/4.9 after that i plugged it into the same equation but for the x components and got as far as: 201.24=V_{i}cos35(V_{i}sin35/4.9) what do I do after that to solve for V_{i}? 


#2
Dec208, 06:14 PM

HW Helper
P: 4,433

In the projectile chapter, you will find three expressions. One for maximum height, second for time flight and third for range of the projectile. Use the third formula to find the initial velocity.



#3
Dec208, 06:16 PM

P: 5

..which projectile chapter?



#4
Dec208, 06:34 PM

HW Helper
P: 4,433

How to find initial velocity in projectile motion problem..?
In any physics text book can you find the equations of projectile motion?



#5
Dec208, 06:38 PM

P: 5

no, my teacher isn't teaching it from a text book. we just learned to use galileo's equations for motion but break them into x and y..
I know for certain that we are expected to use the d=V_{i}t+1/2at^{2} equation first with y components to find t and then plug that in. I'm just not sure what to do once i plugged it in. I guess my question is more related to the trig/algebra involved in finding the solution 


#6
Dec208, 06:52 PM

HW Helper
P: 4,433

Substitute the values of sin35 and cos35 and solve for Vi.



#7
Dec208, 06:55 PM

P: 5

what do you mean substitute the values?



#8
Dec208, 07:01 PM

HW Helper
P: 4,433

201.24=Vicos35(Visin35/4.9) = Vi*2(sin35)(cos35)/4.9. Find the values of sin 35 and cos35 and put it in the above equation to get Vi.



#9
Dec208, 07:07 PM

P: 5

thank you soooo much! I finally got it (:



#10
Oct2509, 09:41 PM

P: 8

is the Vi 14.47?



#11
Oct2509, 10:02 PM

P: 10

Some basic notes
One nice thing about projectile motion is that the horizontal motion and vertical motion are independent of one another. Once thrown, the projectile has constant velocity in the x direction (acceleration in the x direction is zero). Its velocity in the y direction is effected by gravity (if you're using toward earth as your negative y direction, a_y roughly equals 9.8 m/s^2). If you know the angle of the initial velocity, you can find the x and y components by drawing a right triangle with v_0 as the hypotenuse (the magnitude of the velocity will be the length of that side). Sometimes you will know know the magnitude right away, so you need to think about what x and y are in terms of trig functions. The way many simple projectile motion problems are set up, the x component is simply v_0*cos(theta) and the y component is v_0*sin(theta). (It depends on your coordinate system and what angle(s) were given.) 


#12
Nov2309, 06:41 PM

P: 1

Just wondering is the 4.9 in the first equation the vertical distance or how did that get there? I have a similar problem:A cannon is firing a ball at the top of a cliff.
What is the velocity of the cannonball if the horizontal distance between the cannon and the base of the cliff is 1000 meters, the height of the cliff is 60 meters and the cannon makes a 53 degree angle with respect to the horizontal. Thanks 


#13
Nov2309, 06:54 PM

P: 179

[tex]x=v_ocos\theta t[/tex]
[tex]y=y_o + v_ot  \frac {1}{2} g t^2[/tex] Solve the first one for t and substitute into the second equation to find v. 


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