A daredevil on a motorcycle leaves the end of a ramp with a speed of 29.0 m/s as in the figure below. If his speed is 26.7 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance.
Wouldn't it depend on his angle of launch?
Angle of launch is not given in the problem
42
To find the maximum height that the daredevil reaches, we can use the principles of conservation of energy.
The total mechanical energy of the daredevil at any point in the path is the sum of his kinetic energy (KE) and his potential energy (PE). Since we are ignoring friction and air resistance, the total mechanical energy is conserved.
At the start of the ramp, the total mechanical energy consists only of kinetic energy:
KE1 = (1/2) * m * v1^2
where m is the mass of the daredevil and v1 is his speed at the start of the ramp.
At the peak of the path, the kinetic energy is reduced to zero, and the total mechanical energy is solely potential energy:
PE2 = m * g * h
where g is the acceleration due to gravity and h is the maximum height reached.
Equating the mechanical energies at the start and the peak, we have:
KE1 = PE2
(1/2) * m * v1^2 = m * g * h
We can cancel out the mass (m) from both sides of the equation, yielding:
(1/2) * v1^2 = g * h
We are given v1 = 29.0 m/s. To find g, we can use the approximate value of 9.8 m/s^2.
Substituting the given values into the equation, we have:
(1/2) * (29.0 m/s)^2 = (9.8 m/s^2) * h
Simplifying the equation gives us:
(1/2) * 29.0^2 = 9.8 * h
Solving for h:
h = (1/2) * (29.0^2) / 9.8
h ≈ 43.9 meters
Therefore, the maximum height reached by the daredevil is approximately 43.9 meters.