A daredevil leaves the end of a ramp with a speed of 32.0 m/s if his speed is 29.8 m/s when he reaches the peak of the path, what is the maximum height that he reaches?
how do i do thiss?
well,his horzontal speed is all he has at the peak, so vertical speed at launch is
vi^2=32^2-29.8^2
solve for Vi
Then for height,
Vi^2=2*g*h solve for height
vi=11.66
h=2.124
To find the maximum height reached by the daredevil, we can use the principle of conservation of energy. The total energy of the daredevil at any point on the path will be the sum of his kinetic energy (KE) and his potential energy (PE).
At the starting point on the ramp, the energy is in the form of kinetic energy only, since there is no change in height yet. As the daredevil reaches the peak of the path, his speed decreases to 29.8 m/s. At this point, all of his initial kinetic energy will be converted into potential energy.
The equations for kinetic energy (KE) and potential energy (PE) are as follows:
KE = (1/2) * m * v^2
PE = m * g * h
Here, m represents the daredevil's mass, v is the velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the maximum height reached.
Since the daredevil's mass is common in both equations, we can cancel it out. The initial kinetic energy (KE_i) will be equal to the potential energy at the peak (PE_f):
(1/2) * m * v_i^2 = m * g * h
Simplifying the equation by canceling out mass:
(1/2) * v_i^2 = g * h
Now we can solve for the maximum height (h) using the given values:
(1/2) * (32.0 m/s)^2 = 9.8 m/s^2 * h
h = (1/2) * (32.0 m/s)^2 / 9.8 m/s^2
h ≈ 51.8 meters
Therefore, the daredevil reaches a maximum height of approximately 51.8 meters.