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 18.2 above the river, whereas the opposite side is a mere 1.6above the river. The river itself is a raging torrent 64.0 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). The speed of the car traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side is 34.78 m/s.

(B). The velocity on the other side is 39.17 m/s.

Explanation:

Given that,

Distance of side from the river = 18.2 m

Distance of opposite side from the river = 1.6 m

Width = 64.0

We need to calculate the time of the car in the air along the vertical direction

Using equation of motion

$s-s_{0}=u_{y}t+\dfrac{1}{2}a_{y}t^2$

Here, $a_{y}=g$

$s-s_{0}=0+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2(s-s_{0})}{g}}$

Put the value in the equation

$t=\sqrt{\dfrac{2(18.2-1.6)}{9.8}}$

$t=1.84\ sec$

(A). We need to calculate the speed of the car traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side

Using formula of velocity

$v_{x}=\dfrac{d}{t}$

$v_{x}=\dfrac{64.0}{1.84}$

$v_{x}=34.78\ m/s$

(B). We need to calculate the speed of the car just before it lands safely on the other side

Using equation of motion

$s_{y}=u_{y}t+gt$

Put the value in the equation

$s_{y}=0+9.8\times1.84$

$s_{y}=18.032\ m/s$

We need to calculate the magnitude of the velocity on the other side

Using formula of magnitude of the velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

$v=\sqrt{(34.78)^2+(18.032)^2}$

$v=39.17\ m/s$

Hence, (A). The speed of the car traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side is 34.78 m/s.

(B). The velocity on the other side is 39.17 m/s.

Answer 2

We have that  the speed just clear the river and land safely on the opposite side and the speed just before landing will be

• v=35.417m/s

From the question we are told

• 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 18.2 above the river, whereas the opposite side is a mere 1.6above the river.
• The river itself is a raging torrent 64.0 wide.

Speed of car

Generally the equation for the Time   is mathematically given as

$t=\sqrt{2*d*/g}\\\\Therefore\\\\t=\sqrt{2*16.4/9.8}$

t=1.807

Hence

Speed of travel

$v=\frac{64}{1.8070}\\\\v=35.417m/s$

b)

Since the car has to travel at avg speed of v=35.417m/s

Therefore

the speed of the car just before it lands safely on the other side is

• v=35.417m/s

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