1. The best leaper in the animal kingdom is the puma, which can jump to a height of 12 ft when leaving the ground at an angle of 45°. With what speed, in SI units, must the animal leave the ground to reach that height?
0 = (Vcos45)^2 + 2 (-32) 12
v^2 (1/2) = 768
v^2 = 1536
v = 39.2 ft/s
v = 12 m/s
(Vo sin45)^2 = 2 g H
Vo^2 = 4 g H
Vo = 2 sqrt(g H)
g = 32.2 ft/s^2
Vo = 78.4 ft/s = 53.5 mph
SCB is correct. I made a mistake calculating Vo.
Vo = 39.3 m/s
sin and cos 45 are the same, but in general the sine of the angle from horizontal should be used.
8.49
To determine the speed at which the puma must leave the ground in order to reach a height of 12 ft, we can use the principle of conservation of energy. The initial potential energy when the puma leaves the ground will be equal to the final potential energy at the highest point of the jump.
First, let's convert the height from feet to meters since we want our answer in SI units.
1 foot is equal to 0.3048 meters.
12 ft × 0.3048 m/ft = 3.6576 meters
The potential energy can be calculated using the formula:
Potential energy = m × g × h
Where:
m is the mass of the puma
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height (3.6576 meters)
Since we don't have the mass of the puma, we can cancel out the mass on both sides of the equation by dividing both sides by m:
Potential energy/m = m × g × h / m
Potential energy/m = g × h
Now, we can consider the initial kinetic energy of the puma when it leaves the ground. The kinetic energy can be calculated using the formula:
Kinetic energy = 1/2 × m × v^2
Where:
m is the mass of the puma
v is the velocity of the puma
At the highest point of the jump, the final kinetic energy will be zero since the puma briefly comes to rest. Therefore, the initial kinetic energy will be equal to the potential energy:
1/2 × m × v^2 = g × h
Now, we can solve this equation for v:
v^2 = (2 × g × h) / m
v = √((2 × g × h) / m)
To find the speed at which the puma must leave the ground, we need to know the mass of the puma. Unfortunately, the given information does not provide the mass. Without the mass, we cannot calculate the exact speed in SI units.