A 10000 N car comes to a bridge during a storm and finds the bridge washed out. The 750 N driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.6 m above the river, while the opposite side is a mere 3.60 m above the river. The river itself is a raging torrent 63.0 m wide.
How fast should the car be traveling just as it leaves the cliff in order to clear the river and land safely on the opposite side?
got it....33 m/s
What is the speed of the car just before it lands safely on the other side?
yeah...don't know...not even a little bit
What is the speed of the car? The easy way is to use Energy..
He started with 1/2 m v^2 (and you figured v) plus mgh of Potential energy (h=21.6-3.6). Add those. Then when he landed, he had 1/2 m v^2 of energy, so calcuate v.
The other way is you know the initial horizontal velocity, that is also the final horizontal velocity. The vertical velocity at impact will be v= sqrt(2gh).Add those two velocities as vectors.
To find the speed of the car just before it lands safely on the other side, we can use the principle of conservation of mechanical energy. The car initially has some potential energy due to its height above the river, and as it falls, this potential energy is converted into kinetic energy.
The potential energy (PE) of the car at the starting point is given by PE = m * g * h, where m is the mass of the car, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height difference between the starting point and the landing point.
The kinetic energy (KE) of the car just before it lands on the other side is given by KE = 1/2 * m * v^2, where v is the final velocity of the car.
Since energy is conserved, we can equate the initial potential energy to the final kinetic energy:
PE = KE
m * g * h = 1/2 * m * v^2
The mass of the car cancels out, allowing us to solve for v:
g * h = 1/2 * v^2
v^2 = 2 * g * h
v = sqrt(2 * g * h)
Substituting the given values, we get:
v = sqrt(2 * 9.8 * (21.6 - 3.6)) ≈ 33 m/s
Therefore, the speed of the car just before it lands safely on the other side is approximately 33 m/s.