A projectile is launched with a speed of 35 m/s at an angle of 58° above the horizontal. Find the maximum height reached by the projectile during its flight by using conservation of energy.
m
To find the maximum height reached by the projectile using conservation of energy, we need to consider the initial and final states of the projectile.
Initial state: The projectile is launched with a speed of 35 m/s at an angle of 58° above the horizontal. At this point, the only form of energy possessed by the projectile is its kinetic energy.
Final state: The maximum height reached by the projectile, where its vertical velocity becomes zero. At this point, the only form of energy possessed by the projectile is its potential energy.
Using conservation of energy, we equate the initial kinetic energy to the final potential energy:
1/2 * mv_initial^2 = m * g * h_max
Where:
m = mass of the projectile (not given)
v_initial = initial velocity of the projectile (35 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h_max = maximum height reached by the projectile
Since we are asked to find the maximum height reached, we can solve for h_max:
h_max = (v_initial^2)/(2g)
Substituting the given values:
h_max = (35^2)/(2*9.8)
= 625/19.6
= 31.88 m
Therefore, the maximum height reached by the projectile during its flight, using conservation of energy, is approximately 31.88 meters.