Solving Projectile Motion: Minimum Speed for Daredevil Canyon Jump"

[magi58]

Me and my friend are having a hard time with this question, and it says
"a daredevil jumps a canyon 12 m wide. To do so, he drives a car up a 15 degree incline.

What minimum speed must he achiece to clear the canyon and
what will hi speed be when he reaches the other side?

Can anyone help us? We don't even know how to set up the problem

 

[andrevdh]

"[URL
http://wps.aw.com/aw_young_physics_11

Click on the 3.1 Projectile motion link and then on the
3.1 Solving Projectile Motion Problems (3) link.

Load the simulation by clicking on the 1 icon on the right-hand side of the window that opens. Try and answer the questions.

 

[kyle8921]

I can provide a solution to this problem using the principles of projectile motion and basic equations of physics. First, we need to understand that the motion of the daredevil and the car can be broken down into two components - horizontal and vertical. The horizontal component is affected by the car's velocity, while the vertical component is affected by gravity.

To solve for the minimum speed needed to clear the canyon, we can use the equation: Vmin = √(2gh), where Vmin is the minimum speed, g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the canyon (12 m). Plugging in the values, we get Vmin = √(2*9.8*12) = 15.68 m/s.

To find the speed of the car when it reaches the other side, we can use the equation: Vf = √(Vi^2 + 2ad), where Vf is the final velocity, Vi is the initial velocity (which is the minimum speed needed to clear the canyon), a is the acceleration due to gravity (9.8 m/s^2), and d is the distance traveled horizontally (12 m). Plugging in the values, we get Vf = √(15.68^2 + 2*9.8*12) = 16.41 m/s.

Therefore, the minimum speed needed for the daredevil to clear the canyon is 15.68 m/s and his speed when he reaches the other side will be 16.41 m/s. It is important to note that these calculations assume ideal conditions and do not take into account air resistance, friction, or the weight of the car. In reality, the daredevil may need to achieve a slightly higher speed to account for these factors and ensure a successful jump. I hope this helps and good luck with your calculations!