Projectile Motion from a height

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A projectile is launched at a 35-degree angle from a height of 900 m and lands 2500 m away horizontally. To find the initial velocity, kinematic equations in both x and y directions must be set up, eliminating time from the equations. The equations to use are 2500 = Vocos(35)t for horizontal motion and 900 = (Vosin(35) - 4.9t)t for vertical motion. The second equation forms a quadratic, which can be solved using the quadratic formula to find time. Once time is determined, it can be substituted back to find the initial velocity.
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A projectile is shot at a 35 degree angle from a height of 900 m. The it lands 2500 m away vertically. What was the initial velocity of the launch?



I do not know what relevant equations to use- I've tried using the integrals of position vectors like Vox(t) = 2500 m and Voz(t) - 1/2 gt^2 = 900 m, but I don't know how to solve this without knowing the time of flight.
 
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You can do this without calculus.

HINT:

Use the kinematic equations and set up equations the motion in both the x and y directions. This should give you enough equations to eliminate time from the problem and solve for the initial velocity.
 
Do I use the kinematic equations Vo= at and x= .5Vo(t)?
 
Since I'm given an angle, I feel like I should be using trig. Is there some way I can involve trig in the kinematic equations?
 
OK, I have figured out what I need to do, but need help with the math!

I need to solve for t in the 2 eqs: 2500 = Vocos35t and 900 = (Vosin35 - 4.9t)(t) and then set the equations equal to solve for Vo. But in the second equation, how do I solve for t?!
 
In the second equation you have a quadratic (Bring all terms to one side of the equation). You need to use the quadratic formula to solve for t.
 
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