Answer:
For the purposes of your question, we can think of speed and velocity as being the same thing. Therefore, the kinetic energy of an object is proportional to the square of its velocity (speed). In other words, If there is a twofold increase in speed, the kinetic energy will increase by a factor of four.
Explanation:
During a climb UP the mountain, gravity does NO work on the climber.
Actually, it's more correct to say that gravity does NEGATIVE work
on him. The climber has to DO the positive work to haul himself up.
Work = (mass) x (gravity) x (height) .
For the guy in this problem:
Work = (67 kg) x (9.8 m/s²) x (3,500 meters)
= 2,298,100 joules.
If he eats no candy bars on the way, and completely depends on
his stored body fat for the energy, then he'll burn off
(2,298,100 joules) / (3.8 x 10⁷ joules/kg)
= 0.06 kg of fat.
That's only about 2.1 ounces. We KNOW he'll lose more weight than that,
climbing 11,000 feet. That's because climbing is pretty inefficient.
In addition to the potential energy you have to give your body weight,
you also have to expend energy breathing, digesting, metabolizing,
and sweating.
Answer:
414.9 m
Explanation:
First, become familiar with the horizontal, and vertical vector components.
Vertical vector: Vy = V × sin (θ).
Horizontal vector: Vx = V × cos(θ).
Distance traveled = Velocity vector × time in the air.
Time in the air given Vy = 2 × Vy / g (in respect to the metric of the vector).
Range of the projectile = Vx² / g
Time in the air given Vx = (Vx + √(Vx)² + 2gh) / g.
Given a 28° angle with an initial velocity of 70m/s, we have enough information to calculate!
Vx = 70 m/s × cos(28°) ≈ 61.806 m/s
Vy = 70 m/s × sin(28°) ≈ 32.863 m/s
t = 2 × Vy / g
t = 2 × ≈32.863 / 9.8
t = ≈65.726 / 9.8
t ≈ 6.7 s
Distance traveled (horizontal) = Vx × t = 61.806 × 6.7 ≈ 414.9 m
