Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 36.3 m/s to each pellet. Gun A is fired straight upward, with the pellet going up and then falling back down, eventually hitting the ground beneath the cliff. Gun B is fired straight downward. In the absence of air resistance, how long after pellet B hits the ground does pellet A hit the ground?
A: d=Va*t1-4.9t1^2
B: d=-Va*t2-4.9t2^2
set them equal
-Va*t2-4.9t2^2=Va*t1-4.9t1^2
-Va(t1+t2)=4.9(t2^2-t1^2)
-Va(t1+t2)=4.9 (t2-t1)(t2+t1)
-Va=4.9(t2-t1)
or t1-t2=Va/4.9
the time for the pellets to hit the ground is
A: h + 36.3t - 4.9t^2
B: h - 36.3t - 4.9t^2
I expect you will need to know h (the height of the cliff) to solve for the two values of t and take their difference.
A: t = (-36.3 ±√(1317.69+19.6h))/9.8
B: t = (36.3 ±√(1317.69+19.6h))/9.8
The difference is .20√(1317.69+19.6h) - 7.41
Hmmm. bobpursley is usually right on these matters. I'll have to think on it a bit to see why the initial height makes no difference.
because when the one that was fired upward gets down to cliff level it is traveling downward at the same initial speed down as the first one was.
To find out how long it takes for pellet A to hit the ground after pellet B, we first need to analyze their motion individually.
Let's start with pellet B. Since it is fired straight downward, we assume it accelerates due to gravity. In the absence of air resistance, the acceleration due to gravity is a constant -9.8 m/s² (taking downwards as the positive direction).
We can use the second equation of motion:
s = ut + (1/2)at²
where:
- s is the distance traveled,
- u is the initial velocity,
- t is the time taken, and
- a is the acceleration.
For pellet B, the initial velocity (u) is 36.3 m/s (given) and the acceleration (a) is -9.8 m/s² (due to gravity acting downwards).
Since pellet B is traveling straight downward, the distance traveled (s) will be the height of the cliff, which we do not know yet. Let's call it 'h'. So, the equation becomes:
h = (36.3)t + (1/2)(-9.8)t²
Now, let's move on to pellet A. It is fired straight upward. Again, we assume that it accelerates due to gravity, but this time acting in the opposite direction. So, the acceleration (a) will be +9.8 m/s².
The initial velocity (u) is 36.3 m/s (given), and the distance traveled (s) will still be the height of the cliff, 'h'. Therefore, the equation for pellet A is:
h = (36.3)t + (1/2)(9.8)t²
Since both pellets reach the ground, the value of 'h' for pellet B will be the same as that for pellet A.
Now, we can equate the two equations and solve for the time (t) when both pellets hit the ground:
(36.3)t + (1/2)(-9.8)t² = (36.3)t + (1/2)(9.8)t²
Simplifying the equation:
-4.9t² = 4.9t²
Dividing both sides by 4.9t²:
-1 = 1
This equation does not have a valid solution. Therefore, the pellets will not hit the ground at the same time in the absence of air resistance.
Note: The discrepancy arises due to the different paths taken by the pellets. Pellet A travels a longer distance since it goes up and down, while pellet B falls straight down.