Answer:
v = 15 m / s
Explanation:
In this exercise we are given the position function
x = 5 t²
and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval
let's look for the displacements
t = 0 x₀ = 0 m
t = 3 = 5 3 2
x_{f} = 45 m
we substitute
v = 15 m / s
The mechanical energy for the first and the second ball is
Mass of the first ball = 20 kg
The initial speed at which a cannonball is fired from a cannon =1000 m/s
The angle made by the cannonball while being fired from the cannon = 37°
The maximum height reached by the first ball is,
The maximum height of the first cannonball is 17478.69 m.
The initial speed at which a cannonball is fired from a cannon =1000 m/s
The angle made by the cannonball while being fired from the cannon = 90 °
For the first ball, total mechanical energy= Potential energy at maximum height + kinetic energy at the maximum height
So, the total mechanical energy is,
= mgh \: + \frac{1}{2}mv {}^{2} _{x}[/tex]
Therefore, the total mechanical energy for the first and the
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