Answer:
v₂ = 176.24 m/s
Explanation:
given,
angle of projectile = 45°
speed = v₁ = 150 m/s
for second trail
speed = v₂ = ?
angle of projectile = 37°
maximum height attained formula,
now,
now, equating both the equations
v₂² = 31061.79
v₂ = 176.24 m/s
velocity of projectile would be equal to v₂ = 176.24 m/s
Answer:
Range,
Explanation:
The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.
Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.
So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops
Therefore {tex}R = MV²/2QE{/tex}