The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top
1 answer:
Answer:
v = 44,16 m/s
Explanation:
We will fixate our reference in the starting point from where Dan jumped of, at the top of the Casino. Therefore, the displacement made when dan reached the airbag would be of y= -99,4 m viewed from our reference. We describe the motion of dan with the equation:
Dan jumped from the rest, that means that the initial velocity v_0=0, therefore:
Since Dan is moving in the negative axis regarding our reference point, we take the negative root of the equation.
v_y=-√(2*(-9,81 m/s^2 )*(-99,4 m) )=44,1613 m/s
So, if we don’t take the air resistance into account, Dan would have achieved an velocity of 44,16 m/s when he reached the airbag.
I hope everything was clear with my explanation. If you need anything else, let me know. Have a great day :D
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Answer:
The first part can be solved via conservation of energy.
For the second part,
the free body diagram of the car should be as follows:
- weight in the downwards direction
- normal force of the track to the car in the downwards direction
The total force should be equal to the centripetal force by Newton's Second Law.
where because we are looking for the case where the car loses contact.
Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.
Explanation:
The point that might confuse you in this question is the direction of the normal force at the top of the loop.
We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.