Answer: [tex]V_{0,x } = 297.7 \frac{m}{s}\\V_{0,y } = 637 \frac{m}{s}[/tex]
Explanation:
Hi!
We define the point (0,0) as the intial position of the ball. The initial velocity is [tex](V_{0,x}, V_{0,y})[/tex]
The motion of the ball in the horizontal direction (x) has constant velocity, because there is no force in that direction. :
[tex]x(t) = V_{0,x}t[/tex]
In the vertical direction (y), there is the downward acceleration g of gravity:
[tex]y(t) = -gt^2 + V_{0,y}t[/tex]
(note the minus sign of acceleration, because it points in the negative y-direction)
When the ball hits the ground, at t = 65s, y(t = 65 s) = 0. We use this to find the value of the initial vertical velocity:
[tex]0 = t(-gt + V_{0,y})\\V_{0,y} = gt = 9.8 \frac{m}{s^2} 65 s = 637 \frac{m}{s}[/tex]
We used that g = 9.8 m/s²
To find the horizonttal component we use the angle:
[tex]\tan(65\º) = \frac{V_{y,0}}{V_{x,0}} = 2.14\\V_{x,0} = 297.7\frac{m}{s}[/tex]
