Answer:
The final position made with the vertical is 2.77 m.
Explanation:
Given;
initial velocity of the ball, V = 17 m/s
angle of projection, θ = 30⁰
time of motion, t = 1.3 s
The vertical component of the velocity is calculated as;
[tex]V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s[/tex]
The final position made with the vertical (Yf) after 1.3 seconds is calculated as;
[tex]Y_f = V_yt - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m[/tex]
Therefore, the final position made with the vertical is 2.77 m.