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 !
To find the speed of the car just before it lands safely on the other side, you can follow these steps:
1. Start by analyzing the horizontal motion of the car. The speed of the car in this direction is constant, denoted by U (the initial speed). The distance the car needs to cover horizontally is 51.4 meters.
2. Now, analyze the vertical motion of the car. The car needs to fall down from a height of 16.6 meters to a height of 7.9 meters. Assuming the initial speed down is 0, you can use the equation of motion: distance = (1/2) * acceleration * time^2. In this case, the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2.
3. Applying the equation, we have 8.7 = (1/2) * 9.8 * t^2, where t represents the time it takes for the car to fall.
4. Solving for t, we find that t^2 = 1.77, which means t ≈ 1.33 seconds.
5. Since the time it takes for the car to fall is the same as the time it takes to reach the other side, we can use the horizontal distance and time to find the initial speed. U = distance / time = 51.4 meters / 1.33 seconds = 38.6 meters/second.
Therefore, the initial speed of the car just before it lands safely on the other side would be approximately 38.6 meters/second.