[TSgt]

600ft at 40mph

1 mile = 5280 feet so he went .11 miles

Time = Distance ÷ Speed

.16 seconds?

How’d I do?

[Kartographer]

Tank of gas $60.00
Three day pass to the Grand Canyon National Park $25.00
Going out like Thelma and Louise PRICELESS!

[03A3]

What about acceleration due to gravity and the effects of drag due to wind over a solid object?

[CougarGA7]

Your both wrong. His downward velocity when he went over the edge was 0. He would then begin accelerating downward at a rate of 32ft per second squared.

[Hegewisch Dupa]

His 40 mph was just in the horizontal vector - by the way you calced it he would have hit the ground three times faster if he was driving 120 mph, and that’s just not the case

[Lonesome in Massachussets]

How’d I do?

You fail. His initial velocity would have been horizontal, the descent is vertical. Assuming he accelerated at 32.2 ft/sec/sec in the vertical, starting with no initial vertical velocity he would go splat in squareroot( 2 x 600 / 32.2) = 6.1 seconds. In that 6.1 seconds, again assuming no horizontal acceleration he would have gone about 358 feet in the horizontal.

[calex59]

Yes, but a car doesn’t fall at the speed it was traveling, it falls at a speed of 32 ft per second/per second. The 40 MPH quits as soon as the wheels leave the ground and gravity takes over! However, all things considered you are probably pretty close to the mark, and in the long run it doesn’t really matter.

[antiRepublicrat]

You’re not counting gravity, and that 40 mph was the horizontal vector, not vertical.

I myself won’t be counting wind resistance, launch angle and terminal velocity, but:

It will take 6.1 seconds to land, and he should land 358 feet out while going 134 miles per hour vertically.

I knew physics class would be useful somehow. :)

[jdub]

600ft at 40mph 1 mile = 5280 feet so he went .11 miles Time = Distance ÷ Speed .16 seconds? How’d I do?

Not so good. he was travelling horizontally at 45 mph, theoretically. Adjusting for drag he was probably doing 40 when he hit. But he started at the top at 0 vertically, and accelerated at 32 feet per second, per second.

[r9etb]

How’d I do?

Terrible. The 40 mph governs how far out in the canyon he will travel over the time taken to fall 600 feet.

The time required to drop 600 ft. is driven by gravity, coupled with the initial downward speed of the car.

From your freshman physics class, you will recall that

d = d₀ + v₀t + 1/2 a_grav t²

Since he drove off the edge horizontally, his initial downward speed v₀ is approximately zero, as is his initial distance, d₀.

Thus,

d = 1/2 a_grav t²

d = 600 ft, and a_grav = 32.174 ft/sec^2.

So we can solve for t to get a "hang time" of about 6.1 seconds.

Over that time he would travel approximately 360 feet out from the edge of the canyon.

[sig226]

Assume that our driver fell exactly 600 feet, and his forward velocity carried him out far enough that he hit nothing on the way down. Since his downward velocity was zero when the fall began, he accelerated for about 5 2/3 seconds before he hit the ground. He was going about 180 feet per second at impact, or 123 mph.

Ouch.

[GreyMountainReagan]

Not too good.

Not to be a geek but there is a horizontal and vertical velocity component to the overall velocity.

Initially the vertical velocity was 0 mph, the horizontal was 40 mph.

As the car falls the horizontal velocity decreased to somewhere near zero and the vertical velocity increases according to the equation Vf = Vo+Ag * deltatime.

The time travelled is....blah, blah..

anyway...

neglecting air friction of course.