Ski Jumper Soars 111.4 km/hr Down Slope

[jap90]

A ski jumper acquires a speed of 111.4 km/hr by racing down a steep hill. He then lifts off into the air from a horizontal ramp. Beyond this ramp, the ground slopes downward at an angle of q = 45°. Assuming the skier is in free-fall motion after he leaves the ramp,at what distance down the slope does the skier land?
Any help would be greatly appreciated.
thanks

 

[mbrmbrg]

Hey jap, welcome aboard!
You might want to take a look at posting rules for this forum.
What ideas do you have about this problem? What do you know? What are you looking for? What equations might you use? Will you need to find any other information along the way?

 

[kyle8921]

I would first like to commend the ski jumper for achieving such an impressive speed of 111.4 km/hr. That is a remarkable feat and requires a lot of skill and practice.

Now, to answer the question, we can use the equations of motion to calculate the distance at which the ski jumper will land. Since the skier is in free-fall motion, we can use the equation d = 1/2 * g * t^2, where d is the distance traveled, g is the acceleration due to gravity (9.8 m/s^2), and t is the time in seconds.

To find the time, we can use the equation v = u + at, where v is the final velocity (which in this case is 0 m/s since the skier is in free-fall), u is the initial velocity (111.4 km/hr = 30.94 m/s), and a is the acceleration due to gravity. Solving for t, we get t = u/a = 30.94/9.8 = 3.15 seconds.

Now, we can plug in this value of t into the first equation to get d = 1/2 * (9.8) * (3.15)^2 = 49 meters. Therefore, the skier will land 49 meters down the slope from the ramp.

It is important to note that this calculation assumes that there is no air resistance and that the skier is in a perfect free-fall motion. In reality, air resistance and other factors may affect the distance at which the skier lands. But based on the given information, this is the estimated distance at which the skier will land.