Ball launched at an angle

[Michels10]

Homework Statement

A ball is launched from the origin at an angle of inclination of 50 degrees above the xy plane. If the ball lands at coordinates (4,7,0), find the initial velocity of the ball.

Homework Equations

x=(v_oCos(50)Cos(theta))t
y=(v_oCos(50)Sin(theta))t

The Attempt at a Solution

Unfortunately, I have the answer, but I am unaware of how it was obtained.

x=.3189v_ot
y=.5581v_ot

z=v_osin(50)t-4.9t^2

z=0 when t=.1563

so v_o=sqrt(4/(.3189)(.1563)) = 8.957m/s

Can anyone give me some insight as to how the x and y was solved for?

 

[rl.bhat]

Hi Michels10, welcome to PF.
x= vo*cosθ*t. So t = x/vo*cosθ...(1)
y = vo*sinθ*t - 1/2*g*t^2...(2)
x and y is given. Angle of projection is given.
Substitute the value of t in eq.(2) and solve for vo

 

[Michels10]

Hi, thank you for the response!

I made an effort to plug in t and I did not receive the correct answer.

7=v_o*sin(50)(4/(v_o*cos(50)))-(1/2)*(9.8)*((4/(V_o*cos(50)))^2

does this look correct?

I plugged it into wolfram and ended up with a result of +-(9.21823 i)

 

[rl.bhat]

an angle of inclination of 50 degrees above the xy plane
The above statement indicates that the angle is above the xy plane, and the ball lands on the xy plane at the co-ordinates (4, 7, 0). So here y = 0 and x = sqrt(4^2 + 7^2).
Now try to solve the problem.