A 10000 car comes to a bridge during a storm and finds the bridge washed out. The 700 driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 24.0 above the river, while the opposite side is a mere 6.80 m above the river. The river itself is a raging torrent 66.0 wide.
-- For this question, I know how to find the initial speed. If it asks, "What is the speed of the car just before it lands safely on the other side?" how do i calculate that because it's not the same as final velocity right?
To calculate the initial speed of the car just before it lands safely on the other side of the river, you can use the concept of conservation of energy. We can assume that there are no external forces acting on the car during the leap, such as air resistance.
The initial potential energy of the car is given by its height above the river, which is 24.0 m. The final potential energy of the car is given by its height on the other side of the river, which is 6.80 m. The difference in potential energy is the work done by gravity, which is converted into kinetic energy:
ΔPE = m * g * Δh = KE
Where:
m = mass of the car = 10000 kg
g = acceleration due to gravity = 9.8 m/s²
Δh = difference in height = 24.0 m - 6.80 m = 17.2 m
KE = kinetic energy
Since kinetic energy is given by the equation:
KE = ½ * m * v²
Where:
v = velocity of the car
We can now calculate the initial velocity (speed) of the car just before it lands. Rearranging the equation, we have:
v² = 2 * ΔPE / m
Plugging in the values, we get:
v² = (2 * 10000 kg * 9.8 m/s² * 17.2 m) / 10000 kg
v² = 33.44 m²/s²
Taking the square root of both sides, we obtain:
v ≈ 5.79 m/s
Therefore, the speed of the car just before it lands safely on the other side is approximately 5.79 m/s.