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 leaping the river with his car. The side of the road the car is on is 21.9 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 58.0 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?
hf=hi+vi*time-4.9t^2
1.8=21.9+vi*t-4.9t^2
and
58=vi*t
solve for t in the second equation
t=58/vi
put that in the first equation, solve for vi
Now, speed just before impact
vertical speed=9.8*t
speedtotal=sqrt(vi^2+verticalspd^2)
Bob seems to be using vi both for vertical and horizontal speed
I think in fact the original vertical speed is zero so
h = Hi + 0 -4.9 t^2
1.8 = 21.9 - 4.9 t^2
t^2 = 4.1
t = 2.03 seconds to fall
now the car has to go horizontal
58 meters in 2.03 seconds
so I will call the constant horizontal speed u
where
u = 58/2.03 = 28.6 m/s
then
speed = sqrt (28.6^2 + v^2)
where v = 0 + 9.81*2.03
v = 19.9 m/s
so
speed at landing = 34.9 m/s
by the way, that car is toast due to the 35 m/s vertical speed at landing
To solve this problem, we can break it down into two parts: the horizontal distance the car needs to cover (the width of the river) and the vertical distance the car needs to clear (the height difference between the two sides). Let's solve it step by step.
Step 1: Finding the time of flight
To determine the time the car spends in the air, we can use the kinematic equation for vertical motion:
𝑦 = 𝑣0𝑦𝑡 + (1/2)𝑎𝑦𝑡^2
Since the initial vertical velocity (𝑣0𝑦) is zero and the acceleration (𝑎𝑦) is due to gravity (-9.8 m/s^2), we can simplify the equation to:
𝑦 = (1/2)𝑎𝑦𝑡^2
where 𝑦 is the vertical displacement and 𝑡 is the time of flight.
Given:
- 𝑦 = 21.9 m (height of the road)
- 𝑡 = ?
Using the given equation, we can solve for 𝑡:
21.9 = (1/2)(-9.8)𝑡^2
Divide both sides by (-9.8)
𝑡^2 = 21.9 / (-4.9)
𝑡^2 = -4.4694
Since time cannot be negative, we ignore the negative sign, giving us:
𝑡 ≈ 2.12 seconds
So, the car spends approximately 2.12 seconds in the air.
Step 2: Finding the required initial horizontal velocity
Now, we can calculate the horizontal velocity (𝑣𝑥) required for the car to clear the river.
Given:
- 𝑑 = 58.0 m (width of the river)
- 𝑡 = 2.12 s (time of flight)
- 𝑣𝑥 = ?
The horizontal distance can be calculated using the equation:
𝑑 = 𝑣𝑥𝑡
Rearranging the equation, we get:
𝑣𝑥 = 𝑑 / 𝑡
Substituting the given values, we have:
𝑣𝑥 = 58.0 m / 2.12 s
𝑣𝑥 ≈ 27.36 m/s
Therefore, the car should be traveling at approximately 27.36 m/s horizontally to just clear the river and land safely on the opposite side.
Step 3: Finding the speed of the car just before it lands
To find the speed of the car just before it lands on the other side, we need to consider the vertical motion.
Using the equation for vertical velocity:
𝑣𝑦 = 𝑣0𝑦 + 𝑎𝑦𝑡
Given:
- 𝑣0𝑦 = 0 (initial vertical velocity)
- 𝑎𝑦 = -9.8 m/s^2 (vertical acceleration due to gravity)
- 𝑡 = 2.12 s (time of flight)
𝑣𝑦 = 0 - 9.8 m/s^2 * 2.12 s
𝑣𝑦 ≈ -20.776 m/s (negative sign indicates downward direction)
To find the speed just before landing, we need to consider the horizontal motion. The horizontal velocity (𝑣𝑥) obtained earlier remains constant throughout the motion.
Using the Pythagorean theorem, we can find the speed (𝑣) just before landing:
𝑣 = sqrt(𝑣𝑥^2 + 𝑣𝑦^2)
𝑣 = sqrt((27.36 m/s)^2 + (-20.776 m/s)^2)
𝑣 ≈ 34.41 m/s
Therefore, the speed of the car just before it lands on the other side is approximately 34.41 m/s.
To determine the speed required for the car to clear the river and land safely on the other side, we can use the principle of conservation of mechanical energy. At the point of takeoff, the car's mechanical energy is purely potential energy, given by the equation:
Potential Energy = m * g * h
Where:
m = mass of the car
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height difference between the takeoff point and landing point
To clear the river, the car's potential energy at takeoff should be equal to the potential energy at the landing point, neglecting any losses due to air resistance or friction. Therefore, we can equate the potential energy expressions at the two points:
m * g * h1 = m * g * h2
Where:
h1 = height of the takeoff point
h2 = height of the landing point
We can rearrange the equation to solve for the initial speed of the car:
v = sqrt(2 * g * (h1 - h2))
Where:
v = initial speed of the car
To calculate the speed of the car just before it lands on the other side, we can use the equation of motion:
v^2 = u^2 + 2 * a * s
Where:
u = initial velocity (v from the first part of the question)
a = acceleration due to gravity (approximately 9.8 m/s^2)
s = distance traveled horizontally (width of the river, 58.0 m)
To find the final velocity (speed just before landing), we set the initial velocity to be the initial speed found previously and solve for v:
v^2 = (sqrt(2 * g * (h1 - h2)))^2 + 2 * a * s
Let's plug in the given values and calculate the results.
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?
Given values:
h1 = 21.9 m
h2 = 1.8 m
g = 9.8 m/s^2
v = sqrt(2 * 9.8 * (21.9 - 1.8))
v = sqrt(2 * 9.8 * 20.1)
v = sqrt(392.04)
Therefore, the car should be traveling at a speed of approximately 19.8 m/s (or 19.8 * 3.6 ≈ 71.3 km/h) at the time it leaves the road 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?
Given values:
s = 58.0 m
v^2 = (19.8^2) + 2 * 9.8 * 58.0
v^2 = 392.04 + 1136.8
v^2 = 1528.84
Taking the square root of both sides, we find:
v = sqrt(1528.84)
v ≈ 39.11 m/s (or 39.11 * 3.6 ≈ 140.8 km/h)
Therefore, the speed of the car just before it lands on the other side is approximately 39.11 m/s (or 39.11 * 3.6 ≈ 140.8 km/h).