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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 19.6 m above the river, whereas the opposite side is a mere 1.4 m above the river. The river itself is a raging torrent 61.0 m wide. 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?
Part B
What is the speed of the car just before it lands safely on the other side?

Respuesta :

Answer:

Part a)

[tex]v = 31.66 m/s[/tex]

Part b)

[tex]v = 36.9 m/s[/tex]

Explanation:

Initial height of the car is 19.6 m

final height of the bridge is 1.4 m

now the vertical displacement of the car is given as

[tex]h = 19.6 - 1.4[/tex]

[tex]h = 18.2 m[/tex]

now the time taken by the car to travel the vertical distance is given as

[tex]y = \frac{1}{2}gt^2[/tex]

[tex]18.2 = \frac{1}{2}(9.81)t^2[/tex]

[tex]t = 1.93 s[/tex]

Part a)

Now the speed of the car so that it will just cross the cliff

[tex]v = \frac{d}{t}[/tex]

[tex]v = \frac{61}{1.93}[/tex]

[tex]v = 31.66 m/s[/tex]

Part b)

velocity gain in y direction by the car

[tex]v_y = \sqrt{2gh}[/tex]

[tex]v_y = \sqrt{2(9.81)(18.2)}[/tex]

[tex]v_y = 18.9 m/s[/tex]

speed in x direction is given as

[tex]v_x = 31.66[/tex]

so net speed of the car is given as

[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

[tex]v = \sqrt{31.66^2 + 18.9^2}[/tex]

[tex]v = 36.9 m/s[/tex]

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