Answer:
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