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?

See prev post

To determine the minimum speed the physics professor daredevil should have at the top of the takeoff ramp to make the landing ramp, we can use the principles of projectile motion.

(a) What should be his minimum speed at the top of the takeoff ramp for him to make the landing ramp?

To solve this problem, we need to analyze the components of motion separately in the vertical and horizontal directions.

1. Vertical motion:
The vertical motion can be modeled as a simple free-fall motion under the influence of gravity. We can use the equation of motion:

Δy = v₀y * t + (1/2) * g * t²,

where Δy is the vertical displacement (25 m), v₀y is the initial vertical velocity (unknown), t is the time of flight, and g is the acceleration due to gravity (approximately 9.8 m/s²).

We can rewrite this equation as:

25 m = v₀y * t - (1/2) * 9.8 m/s² * t².

2. Horizontal motion:
The horizontal motion is unaffected by gravity and can be modeled as uniform motion with a constant velocity. We can use the equation of motion:

Δx = v₀x * t,

where Δx is the horizontal displacement (60 m), v₀x is the initial horizontal velocity (unknown), and t is the time of flight.

Using trigonometry, we can relate the initial velocity components to the initial speed v₀:

v₀x = v₀ * cos(50.0°) (horizontal component)
v₀y = v₀ * sin(50.0°) (vertical component)

Since the horizontal displacement is given as 60 m, we have:

60 m = v₀ * cos(50.0°) * t.

Now we have two equations related to time and we can solve for the unknowns.

By rearranging the horizontal motion equation, we can solve for the time of flight (t):

t = 60 m / (v₀ * cos(50.0°)).

Substituting this expression for t into the vertical motion equation, we can solve for the initial vertical velocity (v₀y):

25 m = (v₀ * sin(50.0°)) * (60 m / (v₀ * cos(50.0°))) - (1/2) * 9.8 m/s² * (60 m / (v₀ * cos(50.0°)))².

Simplifying this equation, we can solve for v₀:

25 m = (v₀ * sin(50.0°)) * (60 m / (v₀ * cos(50.0°))) - (1/2) * 9.8 m/s² * (60 m / (v₀ * cos(50.0°)))².

Simplifying this equation will give us the minimum speed (v₀) required at the top of the takeoff ramp for the daredevil to make the landing ramp.

(b) If his takeoff speed was only half the answer in part (a), where will he land?

If his takeoff speed is only half the minimum speed calculated in part (a), we can use the same principles of projectile motion to determine where he will land. We will use the new takeoff speed, but all other variables, such as the ramp heights and distances, will remain the same.

Follow the same steps outlined in part (a), using half the minimum speed as the new value for v₀. This will give us the new values for the time of flight and horizontal displacement. The landing position can then be calculated by using the new horizontal displacement value.