Answer:
<em>The end of the ramp is 38.416 m high</em>
Explanation:
<u>Horizontal Motion
</u>
When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.
The maximum horizontal distance traveled by the object can be calculated as follows:
![\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20d%3Dv%5Ccdot%5Csqrt%7B%5Cfrac%20%20%7B2h%7D%7Bg%7D%7D)
If the maximum horizontal distance is known, we can solve the above equation for h:
![\displaystyle h=\frac {d^2g}{2v^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h%3D%5Cfrac%20%20%7Bd%5E2g%7D%7B2v%5E2%7D)
The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:
![\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h%3D%5Cfrac%20%20%7B70%5E2%5Ccdot%209.8%7D%7B2%5Ccdot%2025%5E2%7D)
![\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20h%3D%5Cfrac%20%20%7B4900%5Ccdot%209.8%7D%7B2%5Ccdot%20625%7D)
h= 38.416 m
The end of the ramp is 38.416 m high
B, this is because the particles in a solid such as the diamond can not move and even though they are locked into place they still vibrate
Answer:
a)
& ![m_c.v_c=m_b.v_b\times \cos\theta](https://tex.z-dn.net/?f=m_c.v_c%3Dm_b.v_b%5Ctimes%20%5Ccos%5Ctheta)
b) ![v_c=0.0566\ m.s^{-1}](https://tex.z-dn.net/?f=v_c%3D0.0566%5C%20m.s%5E%7B-1%7D)
c) ![p_e=2.9218\ kg.m.s^{-1}](https://tex.z-dn.net/?f=p_e%3D2.9218%5C%20kg.m.s%5E%7B-1%7D)
Explanation:
Given:
mass of the book, ![m_b=1.35\ kg](https://tex.z-dn.net/?f=m_b%3D1.35%5C%20kg)
combined mass of the student and the skateboard, ![m_c=97\ kg](https://tex.z-dn.net/?f=m_c%3D97%5C%20kg)
initial velocity of the book, ![v_b=4.61\ m.s^{-1}](https://tex.z-dn.net/?f=v_b%3D4.61%5C%20m.s%5E%7B-1%7D)
angle of projection of the book from the horizontal, ![\theta=28^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D28%5E%7B%5Ccirc%7D)
a)
velocity of the student before throwing the book:
Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.
![u_c=0\ m.s^{-1}](https://tex.z-dn.net/?f=u_c%3D0%5C%20m.s%5E%7B-1%7D)
where:
initial velocity of the student
velocity of the student after throwing the book:
Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.
![m_c.v_c=m_b.v_b\times \cos\theta](https://tex.z-dn.net/?f=m_c.v_c%3Dm_b.v_b%5Ctimes%20%5Ccos%5Ctheta)
where:
final velcotiy of the student after throwing the book
b)
![m_c.v_c=m_b.v_b\times \cos\theta](https://tex.z-dn.net/?f=m_c.v_c%3Dm_b.v_b%5Ctimes%20%5Ccos%5Ctheta)
![97\times v_c=1.35\times 4.61\cos28](https://tex.z-dn.net/?f=97%5Ctimes%20v_c%3D1.35%5Ctimes%204.61%5Ccos28)
![v_c=0.0566\ m.s^{-1}](https://tex.z-dn.net/?f=v_c%3D0.0566%5C%20m.s%5E%7B-1%7D)
c)
Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.
![p_e=m_b.v_b\sin\theta](https://tex.z-dn.net/?f=p_e%3Dm_b.v_b%5Csin%5Ctheta)
![p_e=1.35\times 4.61\times \sin28^{\circ}](https://tex.z-dn.net/?f=p_e%3D1.35%5Ctimes%204.61%5Ctimes%20%5Csin28%5E%7B%5Ccirc%7D)
![p_e=2.9218\ kg.m.s^{-1}](https://tex.z-dn.net/?f=p_e%3D2.9218%5C%20kg.m.s%5E%7B-1%7D)
The higher you go the more potential energy there is, and the lower it is the more kinetic energy there is, so the more kinetic energy there is the higher the ball will bounce.