A. 4% you divide 2000 to 500 and you get 4 which is your answer
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)
Answer:

Given:
Initial speed (u) = 22 m/s
Final speed (v) = 0 m/s (Rest)
Time taken (t) = 4 seconds
To Find:
Distance travelled by car (s)
Explanation:
From equation of motion of object moving with uniform acceleration in straight line we have:

By substituting value of v, u & t in the equation we get:


Distance travelled by car (s) = 44 m
Explanation:
The first equation of motion in kinematics is given by :
.....(1)
u is initial speed
a is acceleration
v is final speed
t is time
Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.
Answer:
Explanation:
Check attachment for solution