Answer:
1.85 m
Explanation:
We can start by calculating how much time takes the ball to cover the horizontal distance that separates the starting point from the crossbar, which is
d = 24 m
The horizontal velocity of the ball is constant and it is
[tex]v_x = u cos \theta = (17)(cos 53.2^{\circ})=10.2 m/s[/tex]
So the time taken to cover the horizontal distance d is
[tex]t=\frac{d}{v_x}=\frac{24}{10.2}=2.35 s[/tex]
Now we can analyze the vertical motion of the ball. The vertical position of the ball at time t is given by
[tex]y(t) = u_y t - \frac{1}{2}gt^2[/tex]
where
[tex]u_y = u sin \theta = (17)(sin 53.2^{\circ})=13.6 m/s[/tex] is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
Substituting t = 2.35 s, we find the vertical position of the ball when it is passing above the crossbar:
[tex]y=(13.6)(2.35)-\frac{1}{2}(9.8)(2.35)^2=4.90 m[/tex]
And since the height of the crossbar is
h = 3.05 m
The ball passes
[tex]\Delta h = 4.90 - 3.05 = 1.85 m[/tex] above the crossbar.
