In this problem, we will be looking at your friend Max, the motorcycle daredevil.
Max is planning the stunt of a lifetime.
He wants to take his bike off a ramp and jump over this brick wall, which reaches a height
of thirty meters off the ground.
The ramp itself reaches a height of five meters off the ground, with an angle of fifty-five
degrees off the horizontal.
If the wall is forty-five meters away from the edge of the takeoff ramp, and forty-five
meters away from the edge of a landing ramp on the other side with identical dimensions,
and Max leaves the ramp with a velocity of 31.2 meters per second, will he clear the
wall and land safely on the other side?
To answer this, we will certainly need to use some equations from our study of kinematics,
so if you haven’t viewed those tutorials yet, check out my four-part series on kinematics
in the classical physics playlist.
Otherwise, go ahead and do a little math to make sure our friend Max will be alright.
Let’s find out how he did, shall we?
The first thing we have to do is split up the velocity vector into X and Y components.
As he leaves the ramp, this is the velocity vector, and the X component will therefore
be 31.2 cosine fifty-five, which is 17.9 meters per second.
The Y component will be 31.2 sine fifty-five, which is 25.6 meters per second.
Now the easiest thing to do with this will be to find out how long Max is in the air.
From there we can figure out lots of other things.
So let’s take this equation, and we can plug in the information for the Y direction,
since gravity is pulling straight down.
Whatever the initial velocity was in the Y direction, when he gets back to that same
point going downwards, the velocity will be the same as when he started, but negative.
So we can plug in 25.6 for V initial, and negative 25.6 for V final.
We also know the acceleration due to gravity.
So we combine these velocities, divide by the acceleration, and we get 5.22 seconds
for the hang time.
Now that we know the time spent in the air, we can easily find out the distance traveled
in the horizontal direction.
This will just be the X component of the initial velocity, which doesn’t change while he’s
in the air, times the hang time.
That gives us 93.4 meters, so if we assume that there’s no wind resistance, he should
get the distance needed to hit the landing ramp.
But we still have to make sure Max will clear the wall.
What is the maximum height of his trajectory?
Well the maximum height will occur at precisely half the hang time, since it will take the
same amount of time to go up as it does to come down.
That’s 2.61 seconds, so what will his height be at that moment?
While we didn’t have to consider gravity for the X direction, for the Y direction we
do, so let’s use this equation.
We can plug in the initial velocity, the time elapsed to reach the peak, and acceleration
due to gravity.
Then we just simplify a little bit, and we end up with 33.4 meters.
The distance from the top of the ramp to the top of the wall is only twenty-five meters,
so it looks like he will get over the wall with room to spare.
Congratulations Max, you’ve done it again.