A physics professor moonlights as a daredevil jumping things on a motorbike. His next stunt will be
to jump the oldman river on his motorcycle. The ramp is being built 100 m above the river and made
an angle of 50.0
o
relative to the horizontal. The landing ramp will be 60.0 m away and 25.0 m below
the takeoff ramp. You can ignore air resistance, the stunt rider plans on wearing his speedo and not
his hip hop clothes.
(a) What should be his minimum speed at the top of the takeoff ramp for him to make the landing
ramp?
(b) If his takeoff speed was only half the answer in part (a), where will he land?
a) 60 = v cos50 t
-25 = v sin50 - 1/2* 9.8* t^2
Solve the first for t and plug into the other
b) deep doo-doo I would assume.
To find the minimum speed the daredevil must have at the top of the takeoff ramp in order to make the landing ramp, we can use the principles of projectile motion.
Let's break down the problem step by step:
(a) What should be his minimum speed at the top of the takeoff ramp for him to make the landing ramp?
Step 1: Resolve the initial velocity into horizontal and vertical components.
The given angle of 50.0 degrees with respect to the horizontal helps us split the initial velocity into vertical and horizontal components.
Vertical component: v0y = v0 * sin(angle)
Horizontal component: v0x = v0 * cos(angle)
Step 2: Calculate the time of flight.
To find the time it takes for the daredevil to reach the landing ramp, we can use the vertical motion equation:
y = y0 + v0yt - 0.5gt^2
Since the daredevil starts at a height of 100 m and lands at a height of -25 m (below the takeoff ramp), we can substitute these values into the equation:
-25 m = 100 m + v0y * t - 0.5 * g * t^2
Step 3: Calculate the horizontal distance traveled.
The horizontal distance traveled can be found using the horizontal component of the initial velocity and the time of flight:
x = v0x * t
Since the horizontal distance traveled should be 60.0 m, we can substitute this value into the equation:
60.0 m = v0x * t
Step 4: Combine the equations.
Combining the equation from Step 2 and Step 3 will allow us to solve for the minimum speed at the top of the takeoff ramp (v0).
Substituting the value of t from the horizontal motion equation into the vertical motion equation, we have:
-25 m = 100 m + (v0 * sin(angle)) * (60.0 m / (v0 * cos(angle))) - 0.5 * g * (60.0 m / (v0 * cos(angle)))^2
Simplifying and rearranging this equation will give us the minimum speed required at the top of the takeoff ramp to make the landing ramp.
(b) If his takeoff speed was only half the answer in part (a), where will he land?
If the takeoff speed is only half the answer in part (a), then the initial velocity (v0) will be halved. We can use the same equations as in part (a) to determine the new landing position. Substitute the new values into the equations for vertical and horizontal motion, then solve for the new landing position.