Answer :
The figure of the problem is missing: find in attachment.
(a) 1.64 s
The ball follows a projectile motion path. The horizontal displacement is given by
[tex]x(t) = v_0 cos \theta t[/tex]
where
[tex]v_0[/tex] is the initial speed
t is the time
[tex]\theta=32.0^{\circ}[/tex] is the angle below the horizontal
We can rewrite this equation as
[tex]t=\frac{x(t)}{v_0 cos \theta}[/tex] (1)
The vertical displacement instead is given by
[tex]y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2[/tex] (2)
where
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting (1) into (2),
[tex]y(t) = -x(t) tan \theta - \frac{1}{2}gt^2[/tex]
We know that for t = time of flight, the horizontal displacement is
[tex]x(t) =50.8 m[/tex]
We also know that the vertical displacement is
[tex]y(t) = -45 m[/tex]
Substituting everything into the equation, we can find the time of flight:
[tex]\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s[/tex]
(b) 36.5 m/s
We can now find the initial speed directly by using the equation for the horizontal displacement:
[tex]x(t) = v_0 cos \theta t[/tex]
where we have
x = 50.8 m
[tex]\theta=32.0^{\circ}[/tex]
Substituting the time of flight,
t = 1.64 s
We find:
[tex]v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s[/tex]
(c) 47.1 m/s at 48.8 degrees below the horizontal
As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to
[tex]v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s[/tex]
The initial vertical velocity is instead
[tex]u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s[/tex]
And it changes according to the equation
[tex]v_y = u_y -gt[/tex]
So at t = 1.64 s (when the ball hits the ground),
[tex]v_y = -19.3 - (9.8)(1.64)=-35.4 m/s[/tex]
So the impact speed is:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s[/tex]
While the direction is:
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}[/tex]
