Question

Will give brainliest, 30 points! :)

Answer 1

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

$v_y = u_y + at$ (1)

to find the time it takes for the ball to reach the maximum height. Here we have:

$v_y = 0$ at the instant when the ball reaches the maximum height

$a=g=\frac{1}{6}(-9.8)=-1.63 m/s^2$ is the acceleration of gravity on the Mooon

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 35 m/s is the initial speed

$\theta=28^{\circ}$ is the angle of projection

Substituting,

$u_y = (35)(sin 28)=16.4 m/s$

And now we can solve (1) to find the time the ball takes to reach the maximum height:

$t=\frac{v_y-u_y}{a}=\frac{0-16.4}{-1.63}=10.1 s$

And since the motion downward is symmetrical, the total time of flight is just twice this time:

$T=2t=2(10.1)=20.2 s$

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

$u_x = u cos \theta = (35)(cos 28)=30.9 m/s$

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:

$d=v_x t$

And substituting the time of flight,

t = 20.2 s

We find

$d=(30.9)(20.2)=624.2 m$

c) 521 m

In this case, we have to re-do the exercise on the Earth, where the acceleration due to gravity is

$g=-9.8 m/s^2$

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:

$t=\frac{v_y-u_y}{a}=\frac{0-16.4}{-9.8}=1.67 s$

And so the time of flight is

$T=2t=2(1.67)=3.34 s$

Therefore, the horizontal distance travelled is

$d=v_x t=(30.9)(3.34)=103.2 m$

And so, the difference between the distance travelled by the ball on the moon and on the Earth is

$\Delta d = 624.2 -103.2=521 m$