Answer:
2.47 m
Explanation:
Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.
The horizontal velocity of the ball is constant:
[tex]v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s[/tex]
and the time taken to cover the horizontal distance d is
[tex]t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s[/tex]
So this is the time the ball takes to reach the horizontal position of the crossbar.
The vertical position of the ball at time t is given by
[tex]y=u_y t - \frac{1}{2}gt^2[/tex]
where
[tex]u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s[/tex] is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration of gravity
And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:
[tex]y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m[/tex]
The height of the crossbar is h = 3.05 m, so the ball passes
[tex]h' = 5.52- 3.05 = 2.47 m[/tex]
above the crossbar.
