Question

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John Carter, it shows Carter leaping about 100 feet up on Mars. John Carter is an Earth man who has been transported to Mars so his leg muscles have been built to handle Earth's gravity while Mars' gravity is a lot less. On Earth, Michael Jordan had a vertical jump velocity of 16 ft/sec. Suppose John Carter could triple that initial jump velocity due to being on Mars, so his initial velocity would be v0= 48 ft/sec. a.) How high could he jump on Mars? b.) How long could he stay in the air before he hit the ground? c.) The movie shows Carter jumping about 100 ft high. Is that about right by the calculus? d.) What would his speed be when he hit the ground?

Answer 1

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

     $H_{max}=\frac{v^2}{2g}$

     $H_{max} = \frac{(48)^2}{2 \times 12}$

               = 96 ft

b). Time he can stay in the air before hitting the ground is

   $T=\frac{2v}{g}$

  $T=\frac{2 \times 48}{12}$

     = 8 seconds

c).  Considering upward motion as positive direction.

     v = u + at

We find the time taken to reach the maximum height by taking v = 0.

     v = u + at

     0 = 16 + (12) t

     $t=\frac{16}{12}$

        $=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t =  $=\frac{4}{3} \ s$  , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$  feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s