A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 19.5m above the river, while the opposite side is a mere 1.6m above the river. The river itself is a raging torrent 58.0m wide.
[1]How fast should the car be traveling just as it leaves the cliff in order just to clear the river and land safely on the opposite side?
[2]What is the speed of the car just before it lands safely on the other side?
h = 19.5 - 1.6 = 17.9 m.
Dx = 58 m.
1. h = 0.5g*T^2 = 17.9
4.9T^2 = 17.9
T^2 = 3.65
T = 1.91 s. = Fall time.
Dx = Vo*T = 58
Vo * 1.91 = 58
Vo = 58/1.91 = 30.3 m/s
2. V^2 = Vo^2 + 2g.h = 0 + 19.6*17.9 =
350.84
V = 18.7 m/s
To find the answers to these questions, we can apply the principles of projectile motion. We'll need to divide the problem into two parts: finding the initial velocity of the car to clear the river and finding the speed of the car just before it lands on the other side.
[1] Finding the initial velocity:
To clear the river, we can use the projectile motion equation for the vertical motion component:
y = y0 + v0y*t - (1/2)gt^2
Here, y is the vertical displacement (difference in height), y0 is the initial height, v0y is the vertical component of the initial velocity, t is the time of flight, and g is the acceleration due to gravity. In our case, the initial height is 19.5m, the final height is 1.6m, and the acceleration due to gravity is 9.8 m/s^2.
We can set up two equations. First, at the time the car reaches its maximum height, its vertical velocity is zero, so:
v0y = gt
Secondly, at the time the car lands on the other side, the vertical displacement is the difference in height:
y = y0 - h
where h is the height of the cliff from the bottom of the river.
Substituting the equations:
h = 19.5m - 1.6m = 17.9m
Using the first equation to solve for t:
t = v0y / g
Substituting the value of h in the second equation:
17.9m = (v0y / g) * v0y - (1/2)g * (v0y / g)^2
Simplifying the equation:
17.9m = v0y^2 / g - (1/2)(v0y^2 / g^2)
Multiplying by 2g^2:
2g^2 * 17.9m = v0y^2 - v0y^2 / g
Rearranging and factoring:
v0y^2 = 2g^2 * 17.9m
v0y = sqrt(2g^2 * 17.9m)
Substituting the values:
v0y = sqrt(2 * 9.8 m/s^2 * 17.9m)
Calculating the value:
v0y = 18.8 m/s
Therefore, the car should be traveling at a vertical velocity of approximately 18.8 m/s just as it leaves the cliff to clear the river.
[2] Finding the speed just before landing:
To find the speed of the car just before it lands on the other side, we need to use the horizontal motion component. Since there is no acceleration horizontally, the horizontal speed of the car is constant throughout the motion.
The horizontal displacement is given as the width of the river, which is 58.0m. The time taken to cross the river horizontally can be found by dividing the horizontal displacement by the horizontal component of the initial velocity:
t_horizontal = horizontal displacement / (v0x)
The horizontal component of the initial velocity, v0x, remains constant throughout the motion. Therefore, we can use the time taken to cross the river horizontally to find the speed of the car just before landing:
v_final = v0x
Substituting the values:
v_final = (58.0m) / (t_horizontal)
Now, let's calculate the value of t_horizontal:
t_horizontal = (58.0m) / (v0x / g)
Since v0x = constant = v0x / g
t_horizontal = (58.0m) / (v0x / g)
t_horizontal = (58.0m) * (g / v0x)
Substituting the values:
t_horizontal = (58.0m) * (9.8 m/s^2 / v0x)
Substituting back into the equation for v_final:
v_final = v0x = (58.0m) / [(58.0m) * (9.8 m/s^2 / v0x)]
Cancelling out the units:
v_final = (1 / 9.8) * v0x
v_final = v0x / 9.8
Therefore, the speed of the car just before it lands safely on the other side is approximately one-ninth (1/9) of the horizontal component of its initial speed.
To find the required speed of the car in order to clear the river and land safely on the other side, we can use the principles of projectile motion. To solve this problem, we can use the equations of motion and the conservation of energy.
Step 1: Calculate the time it takes for the car to travel horizontally across the river.
The horizontal distance the car needs to cover is the width of the river, which is 58.0m.
We can use the equation:
Horizontal distance = Horizontal velocity * Time
To find the horizontal velocity, we need to consider that at the maximum height of the jump, the vertical component of the car's velocity is zero. This means that only horizontal velocity contributes to the distance covered.
Hence, the horizontal velocity remains constant throughout the jump.
Since the horizontal distance is equal to the horizontal velocity multiplied by time, we have:
58.0m = Horizontal velocity * Time
Step 2: Calculate the required vertical velocity to clear the river.
We know that the vertical distance the car needs to clear is the difference in height between the starting position and the ending position, which is 19.5m - 1.6m = 17.9m.
Using the conservation of energy, we can equate the potential energy at the cliff edge (mgh1) to the kinetic energy just before landing at the opposite side (1/2 * mv2):
mgh1 = 1/2 * mv^2
where m is the mass of the car, g is the acceleration due to gravity (9.8 m/s²), h1 is the initial height, and v is the velocity of the car.
Step 3: Find the time it takes for the car to reach the opposite side.
We can use the formula for vertical motion under constant acceleration:
h = h0 + v0t + (1/2)gt²
Since the initial vertical velocity is zero (as the car starts from rest and leaves the cliff edge vertically), we have:
17.9m = 0 + (1/2)gt²
Step 4: Solve the equations.
Solving the equations simultaneously will give us the required speed of the car at the two given points in the problem.
First, solve Step 3 for t:
17.9m = (1/2)gt²
t² = (2 * 17.9m) / g
t ≈ √{ (2 * 17.9m) / g }
Then, substitute this value of t into the equation in Step 1:
58.0m = Horizontal velocity * t
Now solve for the horizontal velocity:
Horizontal velocity ≈ 58.0m / t
Finally, substitute this value of the horizontal velocity into the equation in Step 2:
mgh1 = 1/2 * mv²
Simplifying this equation, we can cancel out the mass (m) on both sides, and solving for v will give us the required velocity.
Using the derived equation at the potential energy level:
v = √{ 2gh1 }
where g is the acceleration due to gravity (approximately 9.8 m/s²) and h1 is the initial height (19.5m).
Using these steps, we can now calculate the required speed of the car.
To find the speed of the car just before it lands safely on the other side, we assume the height is negligible compared to the starting height and use the formula:
v = √{ 2gh }
where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height of the landing side (1.6m).