a) 20.2 s
To find the time of flight of the ball, we just need to study its vertical motion.
The vertical motion of the ball is a uniform accelerated motion (free-fall), so we can use the suvat equation
[tex]v_y = u_y + at[/tex] (1)
to find the time it takes for the ball to reach the maximum height. Here we have:
[tex]v_y = 0[/tex] at the instant when the ball reaches the maximum height
[tex]a=g=\frac{1}{6}(-9.8)=-1.63 m/s^2[/tex] is the acceleration of gravity on the Mooon
[tex]u_y = u sin \theta[/tex] is the initial vertical velocity of the ball, where
u = 35 m/s is the initial speed
[tex]\theta=28^{\circ}[/tex] is the angle of projection
Substituting,
[tex]u_y = (35)(sin 28)=16.4 m/s[/tex]
And now we can solve (1) to find the time the ball takes to reach the maximum height:
[tex]t=\frac{v_y-u_y}{a}=\frac{0-16.4}{-1.63}=10.1 s[/tex]
And since the motion downward is symmetrical, the total time of flight is just twice this time:
[tex]T=2t=2(10.1)=20.2 s[/tex]
b) 624.2 m
The horizontal distance travelled by the ball is completely determined by the horizontal motion, which is a uniform motion at constant speed.
The horizontal velocity of the ball is
[tex]u_x = u cos \theta = (35)(cos 28)=30.9 m/s[/tex]
And it is constant during the entire motion, as there are no forces acting along this direction.
Therefore, the horizontal distance travelled is given by:
[tex]d=v_x t[/tex]
And substituting the time of flight,
t = 20.2 s
We find
[tex]d=(30.9)(20.2)=624.2 m[/tex]
c) 521 m
In this case, we have to re-do the exercise on the Earth, where the acceleration due to gravity is
[tex]g=-9.8 m/s^2[/tex]
The initial velocities along the horizontal and vertical direction are the same, so the time taken for the ball to reach the maximum height is:
[tex]t=\frac{v_y-u_y}{a}=\frac{0-16.4}{-9.8}=1.67 s[/tex]
And so the time of flight is
[tex]T=2t=2(1.67)=3.34 s[/tex]
Therefore, the horizontal distance travelled is
[tex]d=v_x t=(30.9)(3.34)=103.2 m[/tex]
And so, the difference between the distance travelled by the ball on the moon and on the Earth is
[tex]\Delta d = 624.2 -103.2=521 m[/tex]
