Question

Bond....James Bond is out in his specially equipped BMW when he is attacked by “Bad Guys”. Traveling at 220 km/h James realizes that he is about to plunge over a cliff so he presses the Big Red Button and ejects. The seat is propelled upward at an angle of 34 degrees to the horizontal and a speed of 197 km/h , relative to the car. If the cliff is 53 m high: a. How long is James in the air? b. If the river is 1.00 km wide does James land in the water or in the trees on the other side?? (Choose the x-axis and the y-axis as the horizontal and vertical axis respectively)

Answer 1

The velocity of car with respect to the ground is

                                                       $\vec{v}_{c,g}= 220 \,\hat{i} \,\, km/hr = 61.1 \,\hat{i} \,\, m/s$

The velocity of bond with respect to the car is

                                                       $\vec{v}_{b,c}= 197\, cos37^{\circ} \,\hat{i} \,\, km/hr +197\, sin\, 37^{\circ} \,\hat{j} \,\, km/hr$

                                                               $= 157.33 \,\hat{i} \,\, km/hr +118.6 \,\hat{j} \,\, km/hr$

So, the velocity of bond with respect to the ground is                                                                                                                                                                                                              $\vec{v}_{b,g}= \vec{v}_{b,g}+ \vec{v}_{c,g}=\left ( 157.33 \,\hat{i} \,\, km/hr +118.6 \,\hat{j} \,\, km/hr \right ) + \left ( 220 \,\hat{i} \,\, km/hr \right )$

                                                     $=377.33 \,\hat{i} \,\, km/hr +118.6 \,\hat{j} \,\, km/hr$

So, the speed of launch of bond with respect to the ground is

$v_{c,g}= \left | \vec{v}_{c,g} \right | = \sqrt{(377.33)^{2}+(118.6)^{2}}= 395.5\,\, km/hr = 109.87\,\, m/s$

And the angle of launch of bond is

                     $\theta = tan^{-1}\left ( \frac{118.6}{377.33} \right )= 17.44^{\circ}$

Now this is the case of projectile motion.

We know the equation of motion,

                                   $S = ut + \frac{1}{2} at^{2}$

Where, S is displacement, u is initial velocity, a is acceleration of the object, and t is time of motion.

In vertical (y) direction:

$S= \Delta y = 0-53= -53\,\, m$

                    $a_{y} = -9.8 \,\, m/s^{2}$     is acceleration due to gravity

                   $u_{y} = 118.6 \,\, km/hr = 32.93 \,\, m/s$

(a) Putting all values in the equation of motion, we get  

                                  $-53 = 32.93t + \frac{1}{2}\times (-9.8) \times t^{2}$

                          or,    $4.9t^{2}- 32.93 t -53 = 0$

                          or,     $t = \frac{-(-32.93) \pm \sqrt{(-32.93)^{2} - 4 \times 4.9 \times (-53)}}{2 \times 4.9}$

                          or,    $t = \frac{32.93 \pm 46.08}{9.8}$

but time can't be negative. So,

                                 $t =8.06 \,\, s$

Hence, James is in the air for 8.06 seconds.

(b) In horizontal (x) direction:

                            $u_{x} = 377.33 \, \. km/hr = 104.8 \,\, m/s$

                           $a_{x} = 0 \,\, m/s^{2}$

And the horizontal distance travelled by Bond in 8.06 second is

                                  $R = u_{x} t =104.8\times 8.06=845.07\,\, m=0.85\,\, km < 1\,\,km$

Hence, the Bond will land in the river.

For any doubt please comment.