A car travelling at V m/s drives off a 3.0m high horizontal ramp...?

A car travelling at V m/s drives off a 3.0m high horizontal ramp in order to jump across a 10m river. Ignore the effects of air resistance in this problem. The mass of the car is 1200 kg and the mass of the driver is 80kg. 

a) Calculate the minimum speed that the car should leave the ramp in order to land on the road across the river?

b) Assume the car leaves the ramp with a speed of 13m/s. What is the magnitude of the velocity of the car as it impacts the road on the other side of the river?

2 Answers

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  • oubaas
    Lv 7
    1 month ago
    Favorite Answer

    falling time t = √2h/g = √2*3/9.80 = 0.782 sec 

    Vmin = d/t = 10/0.782 = 12.8 m/sec

    Conservation of energy : 

    Vimp. = √ V+2gh = √ 13^2+19.6*3 = √ 169+58.8 = 15.1 m/sec 

  • NCS
    Lv 7
    1 month ago

    a) vertically we have y = ½gt² → → t = √(2y / g)

    t = √(2*3.0m / 9.8m/s²) = 0.78 s

    and so the required velocity is v = d/t = 10m / 0.78s ≈ 13 m/s

    b) v² = u² + 2gh = (13² + 2*9.8*3.0) m²/s² = 228 m²/s²

    v = 15 m/s

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