If a particle is projected at an angle of 30° to the horizontal with kinetic energy E then its kinetic energy at the topmost point would be?
the horizontal velocity is the total velocity multiplied by cos(30º)
at the topmost point, there is no vertical velocity
... so the K.E. is ... E cos²(30º)
To find the kinetic energy at the topmost point of the particle's trajectory, we need to consider the conservation of energy.
At the topmost point, the particle reaches its maximum height and momentarily stops before falling back down. At this point, all of the particle's initial kinetic energy is converted into potential energy.
The initial kinetic energy of the particle is given as E. To find the kinetic energy at the topmost point, we can assume that there is no energy lost due to any external factors such as air resistance. Therefore, the initial kinetic energy E is equal to the potential energy at the topmost point.
The formula for potential energy (PE) is given by:
PE = m * g * h
Where:
m is the mass of the particle,
g is the acceleration due to gravity (approximately 9.8 m/s^2),
h is the height (maximum height reached by the particle).
Since the particle momentarily stops at the topmost point, its final velocity is zero. We can also assume that the mass of the particle remains constant.
Using the conservation of energy, the initial kinetic energy E is equal to the potential energy at the topmost point:
E = PE = m * g * h
Rearranging the equation, we have:
h = E / (m * g)
To find the kinetic energy at the topmost point, we need to subtract the potential energy from the total mechanical energy, which is the sum of kinetic and potential energy.
The kinetic energy at the topmost point is given by:
KE_top = E - PE = E - m * g * h
Substituting the value of h from the previous equation, we get:
KE_top = E - m * g * (E / (m * g))
Simplifying the equation, we find:
KE_top = E - E = 0
Therefore, the kinetic energy at the topmost point is zero.