1. An impala is an African antelope capable of a remarkable vertical leap. In on

Question

1. An impala is an African antelope capable of a remarkable vertical leap. In one recorded leap, a 45 kg impala went into a deep crouch, pushed straight up for 0.21 s, and reached a height of 2.5 m above the ground.

Part A: To achieve this vertical leap, with what force did the impala push down on the ground? (Fnet = ?)

Part B: What s the ratio of this force to the antelope's weight? (Fnet/w)

2. Running on a treadmill is slightly easier than running outside because there is no drag force to work against. Suppose a 60 kg runner completes a 5.0 km race in 16 minutes.

Part A: I solved this. I got D = 5.86 N

Part B: What is this force as a fraction of the runner's weight? ( D/W is the form it should be in but I keep getting this wrong).

3. It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration isn't the engine; it's the tires.

Part A: For typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 80 mph? Suppose that Us = 1.00, Uk = 0.80, and Ur = 0.02. (Solving for delta T).

4. The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver was achieved in a jump from a height of 39,000 m. At this elevation, the density of the air is only 4.3% of the surface density. Estimate the terminal speed f a skydiver at this elevation. Suppose the density of air at sea level is 1.2 kg/m^3, the mass of the skydiver is 75 kg, and the cross section area of the skydiver is 0.72 m^2. (solving for Vterm).

I know there's a lot of questions, but any help would be appreciated. Please show how you did it :)

Explanation / Answer

From energy conservation principle (1/2mV^2 = mgh) we learn that :
V = ?2gh = ?19.6*2.5 = 7.00 m/sec
Now impulse I will apply:
F*t = m*V
F = m*V/t = 45*7/0.21 = 1,500.0 N

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