A car is traveling on a level horizontal rode 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 of the road the car is on 21.3m above the river, while the opposite side is mere 1.8m above the river. The rive itself is a ranging torrent 61.0m wide. How fast should the car be traveling at the time it leaves the road in order just to clear the river and the land safely on the opposite side?. Try the suggestions below or type a new query above. What is the speed of the car just before it lands on the other side?
the time of flight is the time it takes for the car to fall the difference in elevation of the two sides
21.3 - 1.8 = 1/2 * 9.8 * t^2
find t
the car has to travel 61 m in time t
the car's horizontal velocity is constant
its vertical velocity increases due to falling
... Vv = g t = 9.8 * t
speed = √[(horizontal)^2 + (vertical)^2]
To determine the speed of the car just before it lands on the other side, we can use the principles of projectile motion.
First, let's break down the problem and identify the relevant information:
Initial height (h1) = 21.3 m
Final height (h2) = 1.8 m
Distance to be cleared (range) = 61.0 m
Now, we can use the equations of motion to find the velocity of the car just before it lands.
1. First, let's calculate the time it takes for the car to reach the other side. Since the motion is vertical, we can use the following equation:
h2 = h1 + V1⋅t - 0.5⋅g⋅t^2
where V1 is the initial vertical velocity (which we want to find) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Solving for t:
1.8 = 21.3 + V1⋅t - 0.5⋅9.8⋅t^2
Rearranging the equation and factoring:
0.5⋅9.8⋅t^2 - V1⋅t + 19.5 = 0
2. Now, let's determine the time it takes for the car to travel horizontally using the equation:
range = horizontal velocity (Vh) ⋅ t
where Vh is the horizontal velocity. Rearranging the equation:
t = range / Vh
3. Substituting the value of t from equation 2 into equation 1, we get:
0.5⋅9.8⋅(range / Vh)^2 - V1⋅(range / Vh) + 19.5 = 0
This is a quadratic equation in terms of Vh. Solving the quadratic equation will give us two potential values for Vh. The positive value will represent the speed of the car just before it lands on the other side.
Alternatively, you can use a physics simulation software or online tool that can solve projectile motion problems to get a more accurate result.