During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. the driver must get to the other side, so he decides to try leap- ing the river with his car. the side of the road the car is on is 21.3 m above the river, while the opposite side is only 1.8 m above the river. the river itself is a raging torrent 48.0 m wide. (a) how fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) what is the speed of the car just before it lands on the other side?

a) The initial velocity should be 24.0 m/s

b) The final speed is 31.0 m/s

Explanation:

The equations for the position and velocity of the car are as follows:

r = (x0 + v0x · t, y0 + v0y · t +1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = vector position of the car

v0x = initial horizontal velocity

v0y = initial vertical velocity

t = time

y0 = initial vertical position

g = acceleration de to gravity

v = velocity at time t

Please, see the figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the edge of the bridge so that y0 = 0 and x0 = 0.

We know the components of the position vector (r final in the figure) when the car reaches the other side of the river:

r final = (48.0 m, -19.5 m)

Then, using the equations for the x and y-components of the vector r final, we can obtain the initial velocity.

First, let´s find the time it takes the car to reach the other side:

y = y0 + v0y · t +1/2 · g · t²           (y0  and v0y = 0)

y = 1/2 · g · t²

-19.5 m = -1/2 · 9.80 m/s² · t²

-19.5 m · -2 / 9.80 m/s² = t²

t = 2.00 s

Now, we can calculate the initial velocity because we know that at t = 2.00 s the x-component of the vector r final is 48.0 m.

x = x0 + v0x · t      (x0 = 0)

48.0 m = v0x · 2.00 s

v0x = 24.0 m/s

b) When the car lands on the other side, its velocity is given by the velocity vector:

v = (v0x, v0y + g · t)

We already know the x-component of v, so let´s find the y-component:

vy = v0y + g · t       (v0y = 0)

vy = -9.80 m/s² · 2.00 s = -19.6 m/s

The final speed will be the magnitude of the vector v:

v² = ( 24.0 m/s)² + (-19.6 m/s)²

v = 31.0 m/s

answer: a. same unit

explanation:

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.