billybob50
Dec 19, 2011, 05:25 PM
You are riding in a car (moving at a constant velocity) and see an open manhole in the road ahead. You take a cantaloupe out of the grocery bag and hold it out the window of the car. Doing a quick calculation you know how far (in meters) ahead of the hole you have to drop the cantaloupe for it to land in the open hole without bouncing on the road. Choose your own speed of the car (between 10 m/s and 20 m/s) and initial height of the cantaloupe (between 1.0 m and 2.0 m) and show the calculations you used and the resulting lead distance for successfully dropping the fruit.

I'm not asking any one to do it for me I'm asking for help on how to do it because I have no idea how to.

Unknown008
Dec 19, 2011, 11:08 PM
What you basically need to understand is that the cantaloupe will follow a parabolic motion as it is falling.

There is the initial horizontal velocity, that of the car, and a vertical acceleration downwards, that of gravity.

The distance traveled vertically is given by the kinematics equation:

s_v = u_vt + \frac12a_vt^2

The distance travelled horizontally is given by the kinematics equation:

s_h = u_ht + \frac12a_ht^2

As you can see, they appear to be the exact same, except that the initial speeds and accelerations are different.

Indeed, along the horizontal, you only have a constant initial speed uh and no acceleration so that the equation effectively becomes:

s_h = u_ht

And along the vertical, there is no initial speed since the fruit is released and not thrown. So it effectively becomes:

s_v = \frac12a_vt^2

Now, you just need to try in some values for sv and uh to get a value for t which would work in both equations.