Answer:
0.303s
Explanation:
horizontal distance travel = 2.050 m, vertical distance travel = 0.45 m
Using equation of linear motion
Sy = Uy t + 1/2 gt² Uy is the inital vertical component of the velocity, t is the time taken for the vertical motion in seconds, and S is the vertical distance traveled, taken downward vertical motion as negative
-0.45 = 0 - 0.5 × 9.81×t²
0.45 / (0.5 × 9.81) = t²
t = √0.0917 = 0.303 s
<span>Changes in the forms of energy are called "Energy Conversions"
Hope this helps!</span>
Answer:To explian the exact calculation of this pruduct we must not try!
Answer:
See the answers below
Explanation:
We can solve this problem using the principle of energy conservation. That is, the energy is conserved before and after dropping the bag.
For this case we have mechanical energy, which is the sum of the kinetic and potential energies.

where:

Ek = kinetic energy [J] (units of Joules)
Ep = potential energy [J]
In the final Energy (2), there is only potential energy. since when the balloon reaches the maximum height its velocity is zero, that is, there is no kinetic energy.
A)
![m*g*h+\frac{1}{2}*m*v^{2} =m*g*h_{1} \\9.81*50+0.5*(15)^{2}=9.81*h_{1}\\h_{1} = 61.46 [m]](https://tex.z-dn.net/?f=m%2Ag%2Ah%2B%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E%7B2%7D%20%3Dm%2Ag%2Ah_%7B1%7D%20%5C%5C9.81%2A50%2B0.5%2A%2815%29%5E%7B2%7D%3D9.81%2Ah_%7B1%7D%5C%5Ch_%7B1%7D%20%3D%2061.46%20%5Bm%5D)
B)
With the value calculated above we can find the acceleration of the balloon.
The distance traveled is the difference between the maximum height and 50 meters.
![x = 61.46-50\\x = 11.46[m]](https://tex.z-dn.net/?f=x%20%3D%2061.46-50%5C%5Cx%20%3D%2011.46%5Bm%5D)
With the following equation of kinematics.

![0 = 15^{2} +2*a*11.46\\a = - 9.816 [m/s^{2} ]](https://tex.z-dn.net/?f=0%20%3D%2015%5E%7B2%7D%20%2B2%2Aa%2A11.46%5C%5Ca%20%3D%20-%209.816%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign indicates that the acceleration acts downward. That is, in the opposite direction to the movement.
We can use the following equation of kinematics to find the final velocity after 4 seconds.
![v_{f}=v_{o}-a*t\\v_{f}=15-9.816*(4)\\v_{f}=-24.24 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5Cv_%7Bf%7D%3D15-9.816%2A%284%29%5C%5Cv_%7Bf%7D%3D-24.24%20%5Bm%2Fs%5D)
Now the distance:
![v_{f}^{2} =v_{o}^{2}-2*a*x\\(24.24)^{2} =(15)^{2} -2*9.81*x\\x = 18.48 [m]\\x_{f}=50+18.48 = 68.48 [m]](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%20%3Dv_%7Bo%7D%5E%7B2%7D-2%2Aa%2Ax%5C%5C%2824.24%29%5E%7B2%7D%20%3D%2815%29%5E%7B2%7D%20-2%2A9.81%2Ax%5C%5Cx%20%3D%2018.48%20%5Bm%5D%5C%5Cx_%7Bf%7D%3D50%2B18.48%20%3D%2068.48%20%5Bm%5D)
c) Using the following equation of kinematics.
![v_{f}=v_{o}-a*t\\0 = 15-9.81*t\\15=9.81*t\\t = 1.52 [s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3Dv_%7Bo%7D-a%2At%5C%5C0%20%3D%2015-9.81%2At%5C%5C15%3D9.81%2At%5C%5Ct%20%3D%201.52%20%5Bs%5D)
Answer
Explanation:
Convert the time to seconds = 0.6 × 60 × 60
= 2160seconds
Velocity = distance ÷ time
Velocity = 500 ÷ 2160
Velocity = 0.23meters per seconds(m/s)
Acceleration = Velocity ÷ time
Acceleration = 0.23 ÷ 2160
Acceleration = 0.000106meters per seconds ²(m/s²)