[pappyone]

So if I stand on the top of the canyon and shoot a pistol at the bottom, the bullet will only travel 32ft/sec? Just askin, I’m math challenged. I couldn’t add two and two and get 150 billion for a bailout.

[Stat-boy]

The acceleration from gravity is 32ft/second^2. If you drop a bullet, it’s velocity will increase at 32ft/second per each second. This is true for any object until air resistance starts to come into play.

If you shoot a round straight down it’s initial velocity will be so high, say 1500 feet/second. After one second, its velocity would have increased to 1532 feet/second, assuming zero air resistance.

[calex59]

No, because you fired the bullet downward. If you fired the bullet across the canyon it would then be affected by gravity and the downward fall would be 32.2 Ft. Per second Squared. The car would not be traveling vertically at 40 MPH, it would be moving horizontally at 40 MPH and that motion would cease to have a driving effect, in other words the wheels would no longer be supplying motion, once the car was off the edge of the cliff. The 40 MPH would have some effect but the downward speed would be 32.2 Ft per second/per second. I believe it would land about 328 ft. OUT and hit the ground in about 6.1 seconds.

A bullet fired downward is a totally different story. Think things through before you ask questions, most of the time you can figure it out.

[Renegade]

If you shoot it parallel to the bottom of the canyon it will fall at 32 feet/ sec. squared . If you shoot it toward the bottom it will travel at muzzle velocity to the bottom .

[whd23]

So if I stand on the top of the canyon and shoot a pistol at the bottom, the bullet will only travel 32ft/sec? Just askin, I’m math challenged. I couldn’t add two and two and get 150 billion for a bailout.

No. If you shoot the pistol towards the bottom, then the initial vertical velocity of the bullet is not zero. However, it will accelerate at (1/2)*(32 ft/sec^2) due to gravity.

v(t) = (1/2)at^2 + vt

When driving a car off the edge, there is very little initial vertical velocity (depends on the angle of the ground with respect to the drop-off). Assuming the ground is at a 90-degree angle with the drop-off, there will be no initial vertical velocity (v(0) = 0).

[Onelifetogive]

So if I stand on the top of the canyon and shoot a pistol at the bottom, the bullet will only travel 32ft/sec? Just askin, I’m math challenged.

No, it would be traveling the initial speed + 32 ft/second squared - deceleration due to wind resistance. Likely the acceleration due to gravity can be ignored because the gravity component would be neglible compared to the initial speed, and the time would be short.