The potential energy of a projectile at its maximum height is equal to its kinetic energy there . If the velocity of projection is 20m/s its time of flight is ?
To determine the time of flight of a projectile, we can start by using the principle of conservation of mechanical energy.
The potential energy (PE) of a projectile at its maximum height is equal to its kinetic energy (KE) there.
The potential energy at the maximum height is given by the formula:
PE = mgh
Where m is the mass of the projectile, g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height.
The kinetic energy is given by the formula:
KE = (1/2)mv^2
Where m is the mass of the projectile and v is the velocity of projection.
Since the potential energy at the maximum height is equal to the kinetic energy, we can equate the two equations:
mgh = (1/2)mv^2
The mass of the projectile cancels out:
gh = (1/2)v^2
To find the time of flight, we need to know the maximum height. Unfortunately, the given information does not provide the maximum height.
However, we can still solve for the time of flight by using the equation of motion for vertical motion:
h = (1/2)gt^2
Where h is the maximum height, g is the acceleration due to gravity, and t is the time of flight.
Rearranging the equation, we get:
t^2 = (2h) / g
Taking the square root of both sides, we have:
t = √(2h / g)
To solve for the time of flight, we need to know the maximum height, which is not provided in the given information. Therefore, we cannot determine the time of flight based on the given information alone.
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m g h = ½ m v²
h = v² / (2 g)
h = ½ g t² ... v² / (2 g) = ½ g t²
t = v / g .. this is ½ the flight time