Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:






we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:



Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:







t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:



So once we got the two distances we can now find the difference between them:

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
The gravitational force on the car is
(9.8 m/s^2) x (the car's mass in kg).
The unit is newtons.
The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).
There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction:

, where

is the coefficient of friction and

is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:

And so, this is the force that the worker must apply to the crate.