a body is projected upward making angle theta with the particle having velocity 300 metre per second find angle theta show that horizontal range could be maximum and find time of flight and horizontal range
Answers
Answer:Consider an α-particle - the nucleus of a He4 atom - travelling with velocity v and kinetic energy KE along the line connecting its centre with that of a stationary copper atom. As the two positively charged nuclei approach, the electrostatic repulsion force between them slows down the α-particle. The copper nucleus, being massive in comparison, may be assumed to remain stationary, as a first approximation. At some point the kinetic energy will fall to zero, and the total energy of the system will be the potential energy PE of the two charged particles a distance r apart. The nuclei then separate and, as the collision is elastic, retrace their previous trajectories.
From the principle of conservation of energy, KE = PE, and simple expressions for both are available from classical physics.
KE = ½.mα.v²
where mα is the mass of the α-particle, 6.644 x 10^(-27) kg, and
v is its initial velocity, 2.00 x 10^7 m/s.
PE = ke.[q1.q2/r]
where ke is the Coulomb constant, and ke = 1/[4.π.ε] where ε is the
permittivity of a vacuum - a defined value, 8.854 x 10^(-12) F/m.
The charges q1 and q2 of the α-particle and the copper nucleus respectively,
velocity=sqrt(9*10^9*2*29*2.56*10^-38*2/10^-13*4*6.6*10^-27)
=3.16*10^6m/s
Explanation: