Answer:
The ball stops instantaneously at the topmost point of the motion.
Explanation:
Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.
The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.
The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that ball will reach at maximum height at
t = 3 s
now we will have

now we have


Now maximum height above ground is given as



Part b)
Height of the fence is given as



Part c)
As we know that its horizontal distance moved by the ball in 5.5 s is given as



now total time of flight is given as

so range is given as



so the distance from the fence is given as


Answer:
C.<u>ten</u><u> </u><u>times</u><u> </u><u>the</u><u> </u><u>intensity</u><u>.</u>
Answer:
The high of the ramp is 2.81[m]
Explanation:
This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.
If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.
We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.
![E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%5C%5C%5C%5Cwhere%3A%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5Cv%20%3D%202.8%5Bm%2Fs%5D%5C%5Cm%3D%5Cfrac%7BE_%7Bk%7D%7D%7B0.5%2Av%5E%7B2%7D%20%7D%20%5C%5Cm%3D%5Cfrac%7B3.8%7D%7B0.5%2A2.8%5E%7B2%7D%20%7D%20%5C%5Cm%3D0.969%5Bkg%5D)
Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.
![E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\](https://tex.z-dn.net/?f=E_%7Bp%7D%2BW_%7Bf%7D%3DE_%7Bk%7D%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5CW_%7Bf%7D%3D23%5BJ%5D%5C%5CE_%7Bk%7D%3D3.8%5BJ%5D%5C%5C)
And therefore
![m*g*h + W_{f}=3.8\\ 0.969*9.81*h - 23= 3.8\\h = \frac{23+3.8}{0.969*9.81}\\ h = 2.81[m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%20%2B%20W_%7Bf%7D%3D3.8%5C%5C%200.969%2A9.81%2Ah%20-%2023%3D%203.8%5C%5Ch%20%3D%20%5Cfrac%7B23%2B3.8%7D%7B0.969%2A9.81%7D%5C%5C%20h%20%3D%202.81%5Bm%5D)