Answer:
Elevation =31.85[m]
Explanation:
We can solve this problem by using the principle of energy conservation. This consists of transforming kinetic energy into potential energy or vice versa. For this specific case is the transformation of kinetic energy to potential energy.
We need to first identify all the input data, and establish a condition or a point where the potential energy is zero.
The point where the ball is thrown shall be taken as a reference point of potential energy.
![E_{p} = E_{k} \\where:\\E_{p}= potential energy [J]\\ E_{k}= kinetic energy [J]](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3D%20E_%7Bk%7D%20%5C%5Cwhere%3A%5C%5CE_%7Bp%7D%3D%20potential%20energy%20%5BJ%5D%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D)
m = mass of the ball = 300 [gr] = 0.3 [kg]
v = initial velocity = 25 [m/s]
![E_{k}=\frac{1}{2} * m* v^{2} \\E_{k}= \frac{1}{2} * 0.3* (25)^{2} \\E_{k}= 93.75 [J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20m%2A%20v%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D%20%5Cfrac%7B1%7D%7B2%7D%20%2A%200.3%2A%20%2825%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%3D%2093.75%20%5BJ%5D)
![93.75=m*g*h\\where:\\g = gravity = 9.81 [m/s^2]\\h = elevation [m]\\replacing\\h=\frac{E_{k}}{m*g} \\h=\frac{93.75}{.3*9.81} \\h=31.85[m]](https://tex.z-dn.net/?f=93.75%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%20%3D%20gravity%20%3D%209.81%20%5Bm%2Fs%5E2%5D%5C%5Ch%20%3D%20elevation%20%5Bm%5D%5C%5Creplacing%5C%5Ch%3D%5Cfrac%7BE_%7Bk%7D%7D%7Bm%2Ag%7D%20%5C%5Ch%3D%5Cfrac%7B93.75%7D%7B.3%2A9.81%7D%20%5C%5Ch%3D31.85%5Bm%5D)
just analyze it in this way:
20cos30*=10( radical 3 )
20sin30*=10
i think what is needed is for people to see that alt. energy sources are very useful and inexpensive.
D. F=ma
F is for force, and that equals two things, M for mass and A for acceleration. When mass is accelerated, it gives you force. Force equals multiplying mass and its acceleration.