The horizontal distance covered by the ball before hitting the water is 70.4 m
Explanation:
The motion of the ball is the motion of a projectile, so it consists of two independent motions:
We start by calculating the time of flight of the ball. This can be done by analyzing the vertical motion. We can use the following suvat equation:
[tex]s=u_y t + \frac{1}{2}at^2[/tex]
where:
s = -16.5 m is the vertical displacement of the ball (it is negative because we take upward as positive direction)
[tex]u_y[/tex] is the initial vertical velocity of the ball, which is given by
[tex]u_y = u sin \theta[/tex]
where
u = 23.5 m/s is the initial velocity
[tex]\theta=33.5^{\circ}[/tex] is the angle of projection
Substituting,
[tex]u_y=(23.5)(sin 33.5^{\circ})=13.0 m/s[/tex]
[tex]a=g=-9.8 m/s^2[/tex] is the acceleration of gravity, downward
Substituting everything into the equation we get:
[tex]-16.5=13.0t-4.9t^2\\4.9t^2-13.0t-16.5=0[/tex]
Solving the equation for t, we find the time of flight of the ball:
t = -0.94 s
t = 3.59 s
We ignore the 1st solution since it is negative, so the ball reaches the water after 3.59 seconds.
Now we analyze the horizontal motion of the ball. The horizontal velocity is constant and it is:
[tex]v_x=u cos \theta=(23.5)(cos 33.5^{\circ})=19.6 m/s[/tex]
Therefore, the horizontal distance covered in a time t is
[tex]d=v_x t[/tex]
And substituting t = 3.59 s, we find
[tex]d=(19.6)(3.59)=70.4 m[/tex]
So, the horizontal distance covered by the ball before hitting the water is 70.4 m.
Learn more about projectile motion:
brainly.com/question/8751410
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