Question

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 18.9 m m above the river, whereas the opposite side is a mere 2.3 m m above the river. the river itself is a raging torrent 60.0 m m wide.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? B) What is the speed of the car just before it lands safely on the other side?

Answer 1

Answer: 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.

Explanation: NO EXPLANTION

Answer 2

Initial horizontal velocity should be approximately
`32.1\; {\rm m\cdot s^(-1)}`.

Velocity right before landing would be approximately
`37.3\; {\rm m\cdot s^(-1)}`.

(Assumptions: the vehicle from on a flat surface with an initial vertical velocity of
`0`; air resistance is negligible;
`g = 9.81\; {\rm m\cdot s^(-2)}`.)

Step-by-step explanation:

Under the assumptions, acceleration of the vehicle in the vertical direction would be
`a_(y) = g = 9.81\; {\rm m\cdot s^(-2)}`. It is given that the vertical displacement of the vehicle would be
`x_(y) = (18.9 - 2.3)\; {\rm m} = 16.6\; {\rm m}`.

Rearrange the SUVAT equation
`x_(y) = (1/2)\, a_(y)\, t^(2)` to find the time
`t` the vehicle was in the air:

`\begin{aligned}t &= \sqrt{(2\, x_(y))/(a_(y))} \\ &= \sqrt{(2\, (18.9 - 2.3))/(9.81)}\; {\rm s} \approx 1.8397\; {\rm s}\end{aligned}`.

For the vehicle to exactly reach the opposite side, its horizontal displacement would be
`x_(x) = 60.0\; {\rm m}`, same as the width of the river.

Also under the assumption, horizontal velocity of the vehicle would be constant while in the air. To find the horizontal velocity
`v_(x)` of the vehicle, divide the horizontal displacement by time the vehicle in the air:

`\begin{aligned}v_(x) &= (x_(x))/(t) \\ &\approx (60.0)/(1.8397)\; {\rm m\cdot s^(-1)} \\ &\approx 32.62\; {\rm m\cdot s^(-1)}\\ &\approx 32.6\; {\rm m\cdot s^(-1)} \end{aligned}`.

Right before landing, vertical velocity of the vehicle would be increase at a rate of
`a_(y) = 9.81\; {\rm m\cdot s^(-2)}`.

After
`t \approx 1.8397\; {\rm s}` in the air (right before landing) the vertical velocity of the vehicle would be:

`\begin{aligned}v_(y) &= u_(y) + a_(y)\, t \\ &\approx 0\; {\rm m\cdot s^(-1)} + (9.81)\, (1.8397)\; {\rm m\cdot s} \\ &\approx 18.05 \; {\rm m\cdot s^(-1)} \\ &\approx 18.0\; {\rm m\cdot s^(-1)}\end{aligned}`.

Apply the Pythagorean Theorem to find the velocity
`v` of the vehicle right before landing:

`\begin{aligned}v &= \sqrt{{v_(x)}^(2) + {v_(y)}^(2)} \\ &\approx \sqrt{(32.62)^(2) + (18.05)^(2)}\; {\rm m\cdot s^(-1)} \\ &\approx 37.3\; {\rm m\cdot s^(-1)}\end{aligned}`.