# Ball launched at an angle

• Michels10
In summary, the problem involves finding the initial velocity of a ball launched from the origin at a 50 degree angle above the xy plane, given that it lands at coordinates (4,7,0). Using the equations x= vo*cosθ*t and y=vo*sinθ*t-1/2*g*t^2, where t=x/vo*cosθ, we can solve for the initial velocity. Plugging in the given values, we get a result of +-(9.21823 i).

## 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=(voCos(50)Cos(theta))t
y=(voCos(50)Sin(theta))t

## The Attempt at a Solution

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

x=.3189vot
y=.5581vot

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

z=0 when t=.1563

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

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

Last edited:
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

Hi, thank you for the response!

7=vo*sin(50)(4/(vo*cos(50)))-(1/2)*(9.8)*((4/(Vo*cos(50)))^2

does this look correct?

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

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.