A motorcyclist is trying to leap across a canyon by driving horizontally of the cliff that is 70.0 m high at 38.0 m/s. Ignoring air resistance, use energy conservation to find the speed with which the cycle strikes the ground at a height of 35.0m on the other side of the canyon. You do not need to know how wide the canyon is!
net change in energy = 0
(change in PE) + (change in KE) = 0
m g (h - H) + 1/2 m (v^2 - V^2) = 0
where h = 35 m, H = 70 m, V = 38 m/s
Solve for v:
[Notice that the mass, m, cancels in the equation]
v = 46.15 m/s
12
Well, this motorcyclist is really taking a leap of faith, quite literally! So let's calculate the speed with which they strike the ground on the other side.
To do this, we can use the principle of conservation of energy. At the top of the cliff, the motorcyclist only has gravitational potential energy, which can be calculated using the formula PE=mgh, where m is the mass of the motorcyclist, g is the acceleration due to gravity, and h is the height of the cliff.
So, the initial potential energy is PE_initial = m * g * h, where h = 70.0 m.
Now, when the motorcyclist is at a height of 35.0 m on the other side of the canyon, all their initial potential energy has been converted into kinetic energy, which can be calculated using the formula KE = (1/2) * m * v^2, where v is the velocity of the motorcyclist.
Setting the initial potential energy equal to the final kinetic energy, we have:
m * g * h = (1/2) * m * v^2
Since the mass of the motorcyclist cancels out, we can calculate the velocity, v:
g * h = (1/2) * v^2
Substituting the given values, g = 9.8 m/s^2, h = 35.0 m:
(9.8 m/s^2) * (35.0 m) = (1/2) * v^2
Solving for v, we find:
v^2 = (2 * 9.8 m/s^2 * 35.0 m)
v^2 = 686 m^2/s^2
Taking the square root of both sides, we find the speed with which the cycle strikes the ground:
v = √686 m/s ≈ 26.2 m/s
So, according to my calculations, the motorcyclist will hit the ground on the other side of the canyon at a speed of approximately 26.2 m/s. I hope they have a really good landing gear!
To find the speed with which the motorcycle strikes the ground on the other side of the canyon, we can use the principle of conservation of energy.
The potential energy (PE) of the motorcycle initially at a height of 70.0 m is given by:
PE = mgh
where m is the mass of the motorcycle, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from the base.
The kinetic energy (KE) of the motorcycle when it strikes the ground on the other side of the canyon at a height of 35.0 m can be calculated using:
KE = (1/2)mv^2
where v is the velocity of the motorcycle.
Since there is no air resistance, the total mechanical energy (sum of potential energy and kinetic energy) of the system will remain constant throughout the motion. Therefore, we can equate the initial potential energy to the final kinetic energy:
mgh = (1/2)mv^2
Canceling the mass from both sides of the equation, we get:
gh = (1/2)v^2
Now we can substitute the known values into the equation. The acceleration due to gravity is g = 9.8 m/s^2, the initial height is h = 70.0 m, and the final height is h = 35.0 m. Plugging these values in, we have:
(9.8 m/s^2)(70.0 m) = (1/2)v^2
Now we can solve for v. Rearranging the equation, we get:
v^2 = (2)(9.8 m/s^2)(70.0 m)
v^2 = 1372 m^2/s^2
Finally, taking the square root of both sides gives us the speed with which the motorcycle strikes the ground on the other side of the canyon:
v = √1372 m/s
v ≈ 37.06 m/s
Therefore, the speed with which the cycle strikes the ground on the other side of the canyon is approximately 37.06 m/s.