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 23.0 m above the river, whereas the opposite side is a mere 1.5 m above the river. The river itself is a raging torrent 52.0 m wide. You may want to review (Pages 75 - 82) . For help with math skills, you may want to review: Vector Magnitudes For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Different initial and final heights. Part A How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side

So, the car should be travelling at a speed of 24.82 m/s

Explanation:

Using Δy = y₂ - y₁ = v₁t + 1/2at², We find the time it takes the car to land on the other side.

Given y₂ = 1.5 m

(the height of the other side above the river), y₁ = 23.0 m (the height of the car's side above the river), v₁ = initial vertical velocity = 0, a = -g = -9.8 m/s²

Δy = v₁t + 1/2at² = 0 - 1/2gt² = -1/2gt²

t = √-(2Δy/g) =√-(2(y₂ - y₁)/g) = √-(2(1.5 - 23.0)/9.8) = √-2(-21.5)/9.8) = √(43/9.8) = √4.388 = 2.095 s ≅ 2.10 s

We now find the initial horizontal velocity u₁ using Δx = x₂ - x₁ = u₁t + 1/2at²

Given Δx = 52.0 m the width of the river, a = 0 (no horizontal acceleration) and t = 2.095 s

Δx = u₁t + 1/2at² = u₁t + 0 = u₁t

u₁ = Δx/t = 52.0 m/2.095 s = 24.82 m/s

So, the car should be travelling at a speed of 24.82 m/s just as it leaves the cliff in order to just clear the river and land safely on the other side.

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