when i shoot my air rifle vertically in the air it takes 16.5 seconds for the pellet to fall to the ground....

Question

i am shooting a pellet vertically. the whole journey takes 16.5 seconds. how many metres did the pellet reach before it turned round and fell under gravity?

when the pellet turns around it must be stationary and then fall under gravity at 9.8ms^2 so can the height the pellet reached be calculated from only knowing the time for the whole journey?

thanks

Answer 1

huh?

Answer 2

On the way down the pellet will reach terminal velocity which I would guess would be about 150 feet per second, a second guess would be that the pellet takes about 1 third of the time for the ascent making about 10 seconds for the fall, that comes out at around 1500 feet say 450 metres, that is quite a powerful gun.

Answer 3

There's not enough information to solve this problem.

If you are standing on the ground, shooting vertically, the pellet goes up a certain distance, falls that same certain distance, and then keeps falling until it reaches the ground. How far above the ground is the pellet when it begins its journey? Nobody's stating that.

Answer 4

There is enough info here for an estimate. Let´s not concern ourselves with terminal velocity, which is the maximum velocity the bebe can get while accelerating at free fall, and air resistance. It´s a bebe not a spaceshuttle. Let´s set the landing velocity the same as muzzle velocity. This equation does the trick:

d=v0t + (at^2)/2

We start at 0 velocity and 0 time. 0 x 0 is nothing so left is (at^2)/2. A is 9.81m2^2 and t is the time the bebe took to get to its highest point. 8,25 seconds there abouts. The bebe is fired about 2 meters above ground and the time we have is when the bebe falls those extra 2 meters to the ground. Let´s skip that too for now. So:
(9.81x8,25^2)/2 gives 333.8 meters. A rough estimate but gives you an idea. Actual number should be lower. I´d round down to 300 meters.

Answer 5

Not quite. Air retards the pellet both rising and falling.
The initial velocity produced by the air-gun may well be higher than the terminal velocity of the pellet, ( the velocity where air resistance = pellet weight), and the pellet is light enough that terminal velocity may be a factor in the speed of its fall.

Answer 6

btw, if you do shoot exactly vertical, the pellet will fall to the ground some (significant) distance to the west.

This is due to the Coriolis effect (or the rotation of the Earth while the pellet is in the air).

Answer 7

If you ignore air resistance, then yes it can.

d = (1/2)gt^2

Only use half of the 16.5 seconds because the pellet only spends half the time going upward.

So: d = (1/2)(9.8 m/s^2)(8.25 s)^2 = 333.5 m

Of course to figure it out perfectly you would need to involve air resistance and the difference in distance from the end of the barrel from which the pellet leaves and to the ground upon which it returns.

Answer 8

I understand that baring air pressure resistance, when an object reaches it's point of origin (we're assuming that your pointing straight up) it will be travelling at the same speed as it started out at. If that is the case, divide your time in half ( you'll have 8.25 sec.) and use the formula you've stated (I know it as 32 ft per sec. per sec.) you should be able to determine the highest point of the pellet.

Answer 9

Yes = if you assume :-
1) The point of launch is, say, 2m above the ground ...
2) Zero air resistance or at least that effect of air resistance 'on the way up' == same as the effect 'on the way down' (i.e. pellet does not reach terminal velocity on the way down).

Answer 10

The other thing that we would need is the air drag. This is going to be very significant; doing this in a vacuum would simplify matters to the point where the starting height, ending height, and time would be enough, I think.

Answer 11

It depends on the terminal velocity of the pellet as it falls.