No for most students want school to end but mostly they would en in june
Answer:
the ground when the ball hits it
Explanation:
A ball falls freely towards the Earth. If the action force is the Earth pulling down on the ball, then identify which of the following best describe the reaction force:

In physics, motion is the phenomenon in which an object changes its position over time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time. One can also speak of motion of images, shapes, and boundaries.
Movement of any object from one position to another position with respect to the observer is called as Motion. Motion Along a Straight Line: When an object moves along a straight line, the motion of the object is called rectilinear motion. For example; motion of a car on highway.
In physics, motion is the change in position of an object with respect to its surroundings in a given interval of time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, and speed. An object's motion cannot change unless it is acted upon by a force.

Hope It Helps!
<span>ANSWER: the height is same either way and hang time is same either way.
EXPLANATION :
suppose if a basketball player jumps vertically will have a hang time around half a second. The remaining half a second he spends in the top of the jump.</span>
Explanation:
Expression for energy balance is as follows.

or, 
Therefore,


Hence, expression for exit velocity will be as follows.

= ![V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}](https://tex.z-dn.net/?f=V%5E%7B2%7D_%7B1%7D%20%2B%202C_%7Bp%7D%28T_%7B1%7D%20-%20T_%7B2%7D%29%5D%5E%7B0.5%7D)
As
for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.
= ![[(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}](https://tex.z-dn.net/?f=%5B%28350%20m%2Fs%29%5E%7B2%7D%20%2B%202%281.007%20kJ%2Fkg%20K%29%20%2830%20-%2090%29%20K%20%5Cfrac%7B1000%20m%5E%7B2%7D%2Fs%5E%7B2%7D%7D%7B1%20kJ%2Fkg%7D%5D%5E%7B0.5%7D)
= 40.7 m/s
Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.