I have another question for the problem below, so What is the speed of the car just before it lands safely on the other side? would it be the same 38.6 meters/sec

A 10,000N car comes to a bridge during a storm and finds the bridge washed out. The 650-N driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 16.6 m above the river, while the opposite side is a mere 7.90 m above the river. The river itself is a raging torrent 51.4 m wide. 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?

This question has already been posted on here but I need to know how they found the initial velocity.
Physics - Damon, Friday, February 8, 2008 at 5:13pm
Horizontal problem:
Speed = constant = U, the initial speed
distance = U t = 51.4 m
Vertical problem:
falls (16.6 - 7.9) = 8.7 meters DOWN
initial speed down = 0
so
8.7 = (1/2) (9.8) t^2
weight has no effect on this problem, only the acceleration of gravity
8.7 = 4.9 t^2
t^2 = 1.77
t = 1.33 seconds to9 fall
same t to reach the other bank so
U = 51.4 meters / 1.33 seconds
= 38.6 meters/sec
Physics - Charlie, Friday, February 8, 2008 at 5:38pm
Thank you so much.
Physics - Damon, Friday, February 8, 2008 at 6:42pm
You are welcome !

Well, if the car manages to land safely on the other side, it's probably traveling at the speed of "please-don't-crash-please-don't-crash" meters per second. Safety first, after all!

To find the initial velocity of the car just before it lands safely on the other side, you can follow these steps:

1. The horizontal problem states that the speed of the car (U) should be constant.
2. The distance the car needs to cover horizontally is 51.4 meters, which is the width of the river.
3. The vertical problem involves the car falling from a height of 16.6 meters to a height of 7.90 meters.
4. The initial speed (downwards) is 0.
5. Using the kinematic equation, we can find the time it takes for the car to fall vertically.
- The equation is: distance = (1/2) * acceleration * time^2.
- Plugging in the values: 8.7 = (1/2) * 9.8 * t^2.
- Solve for t: t^2 = 1.77.
- Taking the square root, we get t = 1.33 seconds.
6. Since the time it takes to fall vertically is the same as the time to reach the other side horizontally, the time for the car to cross the river is also 1.33 seconds.
7. Finally, divide the horizontal distance by the time: U = 51.4 meters / 1.33 seconds = 38.6 meters/sec.
- This means that the speed of the car just before it lands safely on the other side is 38.6 meters/sec.

To find the speed of the car just before it lands safely on the other side of the river, you can follow these steps:

1. Set up the problem as a horizontal and vertical motion problem.
2. Solve the horizontal problem first. Since there is no horizontal acceleration, the speed of the car remains constant. Let's call this speed U.
3. The distance the car needs to travel horizontally is equal to the width of the river, which is 51.4 meters.
Therefore, U * t = 51.4 meters, where t is the time it takes for the car to travel across the river.
4. Solve the vertical problem next. The car falls vertically from a height of 16.6 meters to 7.9 meters. This is a distance of 8.7 meters downward.
5. The initial speed of the car in the vertical direction is 0.
6. Use the equation for motion under constant acceleration in the vertical direction:
d = (1/2)gt^2, where d is the distance, g is the acceleration due to gravity (9.8 m/s^2), and t is the time.
Plug in the values: 8.7 = (1/2)(9.8)t^2.
7. Solve for t^2: t^2 = 1.77.
8. Take the square root of both sides to find t: t = 1.33 seconds.
9. Since the same time is required to reach the other bank, the time is also 1.33 seconds for the horizontal motion.
10. Divide the distance traveled horizontally by the time: U = 51.4 meters / 1.33 seconds = 38.6 m/s.

Therefore, the car should be traveling at a speed of 38.6 meters per second just as it leaves the cliff in order to clear the river and land safely on the opposite side.