A projectile is launched with a speed of 41 m/s at an angle of 60° above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.
m
Find the vertical velocity (41sin60)
mgh=1/2 m (vvertical)^2 solve for h.
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To find the maximum height reached by the projectile, you can use conservation of energy.
The law of conservation of energy states that the total mechanical energy of a system, which is the sum of kinetic energy (KE) and potential energy (PE), remains constant if no external forces are acting on the system.
At the maximum height, the projectile's velocity is zero, so its kinetic energy is zero. Therefore, all of the initial kinetic energy is transformed into potential energy.
The initial kinetic energy of the projectile can be expressed as:
KE = (1/2) * m * v^2
where m is the mass of the projectile and v is its initial velocity. In this case, the mass (m) is not given, but it cancels out when comparing kinetic and potential energy.
The potential energy at the maximum height can be expressed as:
PE = m * g * h
where g is the acceleration due to gravity and h is the maximum height reached by the projectile.
By equating the initial kinetic energy to the potential energy at the maximum height, we can solve for h.
(1/2) * m * v^2 = m * g * h
Canceling out the mass:
(1/2) * v^2 = g * h
Rearranging the equation:
h = (1/2) * (v^2 / g)
Now, substitute the given values into the equation:
v = 41 m/s
g = 9.8 m/s^2 (approximate acceleration due to gravity on Earth)
h = (1/2) * (41^2 / 9.8)
h ≈ 87.3 meters
Therefore, the maximum height reached by the projectile is approximately 87.3 meters.