The height of the ball would be 4.32 m
Explanation:
Given-
Distance from the ball, s = 17.5 m
Angle of projection, θ = 22.5°
Initial speed, u = 24.5 m/s
Height, h = ?
Let t be the time taken.
Horizontal speed, [tex]u_{x}[/tex] = u cosθ
= 24.5 * cos 22.5°
= 24.5 * 0.924
= 22.64 m/s
Vertical velocity, [tex]u_{y}[/tex] = u sinθ
= 24.5 * sin 22.5°
= 24.5 * 0.383
= 9.38 m/s
We know,
[tex]x = u * cos (theta) * t[/tex]
[tex]17.5 = 22.64 * t\\\\t = 0.77s[/tex]
To calculate the height:
[tex]h = ut - \frac{1}{2}gt^2[/tex]
[tex]h = u sin (theta)t - \frac{1}{2} gt^2[/tex]
[tex]h = 9.38 * 0.77 - \frac{1}{2} * 9.8 * (0.77)^2\\ \\h = 7.22 - 2.90\\\\h = 4.32m\\[/tex]
Therefore, height of the ball would be 4.32 m
