Explanation:
1.
We use the equation
h = , where
h is the height traveled,
g is the acceleration due to gravity and
t is the time taken to reach height h.
We can now calculate t to be
= 0.495 s
Let v be the initial velocity of the player.
The player deaccelarates from v m/s to 0 m/s in 0.495 s at the rate of 9.81 m/s^2.
v = 9.81 m/s^2 x 0.495 s = 4.85 m/s
2.
The player takes 0.3 s to increase his velocity from 0 m/s to 4.85 m/s. So his average accelaration is
4.85 m/s / 0.3 s = 16.2 m/s^2
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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