Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.
At time t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is 7/3 . v1 = 7/3 m/s .
At time t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
v2 = zero .
At time t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is -3/5 . v3 = -0.6 m/s .
<h2>Answer: at an angle

below the inclined plane.
</h2>
If we draw the <u>Free Body Diagram</u> for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight
of the block, which is directly proportional to the gravity acceleration
:

This force is directed vertically at an angle
below the inclined plane, this means it has an X-component and a Y-component:



Therefore the correct option is c
Emily throws the ball at 30 degree below the horizontal
so here the speed is 14 m/s and hence we will find its horizontal and vertical components


vertical distance between them

now we will use kinematics in order to find the time taken by the ball to reach at Allison

here acceleration is due to gravity

now we will have

now solving above quadratic equation we have

now in order to find the horizontal distance where ball will fall is given as

here it shows that horizontal motion is uniform motion and it is not accelerated so we can use distance = speed * time

so the distance at which Allison is standing to catch the ball will be 5.33 m
From the starting depth to the surface, the vertical distance is 35 ft.
From the surface to the peak of the jump, the vertical distance is 27 ft.
From the peak of the jump to the surface, the vertical distance is 27 ft.
From the surface to the ending depth, the vertical distance is 18 ft.
Then the total vertical distance is ...
35 ft + 27 ft + 27 ft + 18 ft = 107 ft