A daredevil decides to jump a canyon of width 10 m. To do so, he drives a motorcycle up an incline sloped at an angle of 15 degrees. What minimum speed must he have in order to clear the canyon?
14m/s
Yes
To find the minimum speed needed for the daredevil to clear the canyon, we can use the principle of conservation of mechanical energy. Let's break down the problem into several steps:
Step 1: Determine the potential energy at the top of the incline.
The potential energy at the top of the incline is given by the formula: PE = mgh, where m is the mass of the daredevil and the motorcycle, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical height of the incline.
Given that the angle of the incline is 15 degrees and the width of the canyon is 10 m, we can use trigonometry to find the height of the incline. The height (h) is given by the formula: h = w * sin(theta), where w is the width of the canyon and theta is the angle of the incline.
So, h = 10 m * sin(15 degrees).
Step 2: Determine the kinetic energy at the bottom of the incline.
The kinetic energy at the bottom of the incline is given by the formula: KE = 0.5 * mv^2, where m is the mass of the daredevil and the motorcycle, and v is the velocity of the motorcycle.
Step 3: Set up the conservation of mechanical energy equation.
The conservation of mechanical energy states that the total mechanical energy at any point must be equal. In this case, the potential energy at the top of the incline is equal to the kinetic energy at the bottom of the incline.
PE = KE
Therefore, mgh = 0.5 * mv^2
Step 4: Solve for the minimum velocity.
Cancel out the mass (m) on both sides of the equation.
gh = 0.5 * v^2
Rearrange the equation to solve for v:
v^2 = 2gh
v = √(2gh)
Now you can substitute the known values into the equation and calculate the minimum velocity needed to clear the canyon.
Assuming the motorcycle acts like a point mass (we don't want him to make it across with just his front wheel!) just drag out your good old equations for a trajectory:
for initial velocity v at angle t, the range is
r = (v^2 sin 2t)/g
10 = (v^2 sin 30°)/9.8
v^2 = 20 * 9.8 = 196
v = 14m/s or 50.4 km/hr
Just to check:
he'll be in the air 10/14 = .71 sec
his vertical velocity starts at 14 sin 15° = 3.62m/s
v = 3.62 - 9.8t
when v=0 (top of the arc) t = .36 sec
just about right. He has to go up and then come back down, taking .72 sec in all.