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 20.5 m above the river, whereas the opposite side is a mere 2.4 m above the river. The river itself is a raging torrent 57.0 m wide.
A) How fast should the car be traveling just as it leaves the cliff in order just to 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) Velocity of car in order just to clear the river and land safely on the opposite side is $29.68 \frac{m}{s}$

(B) The speed of the car is $35.14 \frac{m}{s}$

Explanation:

Given:

Car distance from river $y = 20.5$ m

Mere distance from river $y_{o} = 2.4$ m

River length $x= 57$ m

(A)

For finding the velocity of car,

Using kinematics equation we find velocity of car

  $y - y_{o} = v_{o}t + \frac{1}{2} gt^{2}$

Where $v_{o} = 0$, $g = 9.8$ $\frac{m}{s^{2} }$                                     ( given in question )

$20.5 - 2.4 = \frac{1}{2} \times 9.8 \times t^{2}$

   $t= 1.92$ sec

The speed of the car before it lands safely on the opposite side of the river is given by,

   $v_{x} = \frac{x-x_{o} }{t}$

   $v_{x} = \frac{57-0}{1.92}$                        ( $x_{o} = 0$ )

   $v_{x} = 29.68$ $\frac{m}{s}$

(B)

For finding the speed,

The horizontal distance travel by car,

  $v_{y} = v_{o} + at$

Where $a = 9.8$ $\frac{m}{s^{2} }$

  $v_{y} = 9.8 \times 1.92$

  $v_{y} = 18.81$ $\frac{m}{s}$

For finding the speed,

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

  $v = \sqrt{(29.68)^{2}+ (18.81)^{2} }$

  $v = 35.14$ $\frac{m}{s}$

Therefore, the speed of the car is $35.14 \frac{m}{s}$