There will not be enough momentum from the first hill to cross another hill if he same or larger size because of the way potential energy and kinetic energy works it will not be able go as high as it could go on he fist hill.
Weight can be explained as the force with which the gravity pulls an object. Your weight will not be the same in all planets. In moon, you will weigh far lesser than how much you weigh on the earth. However, in earth and in the moon, your mass will remain the same.
Answer:
The multiverse is a hypothetical group of multiple universes. Together, these universes comprise everything that exists: the entirety of space, time, matter, energy, information, and the physical laws and constants that describe them.
Hope it helped you :D
Answer:
a) t = 3.35[s]; b) t = 1.386[s]
Explanation:
We can solve this problem by dividing it into two parts, for the first 55 [m] and then the second part with the remaining 55 [m].
We will take the initial velocity as zero, as the problem does not mention that the Rock was thrown at initial velocity.
And using kinematics equations:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\where:\\v_{o}=0\\g=gravity = 9.81[m/s^2]\\y=55 [m]\\v_{f}^{2}=0+2*9.81*55\\v_{f}=\sqrt{2*9.81*55} \\v_{f}=32.85[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cwhere%3A%5C%5Cv_%7Bo%7D%3D0%5C%5Cg%3Dgravity%20%3D%209.81%5Bm%2Fs%5E2%5D%5C%5Cy%3D55%20%5Bm%5D%5C%5Cv_%7Bf%7D%5E%7B2%7D%3D0%2B2%2A9.81%2A55%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B2%2A9.81%2A55%7D%20%5C%5Cv_%7Bf%7D%3D32.85%5Bm%2Fs%5D)
Now we can calculate the time:
![v_{f}=v_{o}+g*t\\t=\frac{v_{f}-v_{o}}{g}\\ t=\frac{32.85-0}{9.81}\\ t=3.35[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D%2Bg%2At%5C%5Ct%3D%5Cfrac%7Bv_%7Bf%7D-v_%7Bo%7D%7D%7Bg%7D%5C%5C%20t%3D%5Cfrac%7B32.85-0%7D%7B9.81%7D%5C%5C%20t%3D3.35%5Bs%5D)
Now we can calculate the second time, but using as a initial velocity 32.85[m/s].
The final velocity will be:
![v_{f}^{2}= v_{o}^{2}+2*g*y\\v_{f}=\sqrt{v_{o}^{2}+2*g*y} \\v_{f}=\sqrt{32.85^{2}+2*9.81*55 } \\v_{f}=46.45[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%5C%5Cv_%7Bf%7D%3D%5Csqrt%7Bv_%7Bo%7D%5E%7B2%7D%2B2%2Ag%2Ay%7D%20%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B32.85%5E%7B2%7D%2B2%2A9.81%2A55%20%7D%20%5C%5Cv_%7Bf%7D%3D46.45%5Bm%2Fs%5D)
Now we can calculate the second time:
![t=\frac{46.45-32.85}{9.81} \\t= 1.386[s]](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B46.45-32.85%7D%7B9.81%7D%20%5C%5Ct%3D%201.386%5Bs%5D)
Note: The reason the second time is shorter even though it is the same distance is that the acceleration of gravity increases the speed of the rock more and more as it falls.