Two identical pellet guns are fired simultaneously from the edge of a cliff. These guns impart an initial speed of 27.3 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?
Pellet A:
V = Vo + g*Tr
Tr = -Vo/g = -27.3/-9.8 = 2.79 s. = Rise
time.
Tf1 = Tr = 2.79 s. = Time to Fall back to the edge of the cliff.
Tf2 = Fall time from edge of cliff to
gnd.
Ta = Tr + Tf1 + Tf2 = 2.79 + 2.79 + Tf2 = 5.57 + Tf2. = Time in air, pellet A
Pellet B:
Tb = Tf = Tf2: Same initial velocity and distance.
(5.57+Tf2) - Tf2 = 5.57 s. After pellet
B hits the gnd.
To solve this problem, we can first find the time it takes for pellet B to hit the ground. Since the initial velocity is 27.3 m/s and the acceleration due to gravity is 9.8 m/s² (assuming the downward direction as positive), we can use the kinematic equation:
h = ut + (1/2)at²
Where:
h = height (in this case, the height is the distance from the top of the cliff to the ground)
u = initial velocity
t = time
a = acceleration
For pellet B:
Taking the downward direction as positive, the initial velocity (u) is 27.3 m/s, the acceleration (a) is 9.8 m/s² (due to gravity), and the height (h) is the distance from the top of the cliff to the ground, which we need to find.
Since the pellet is fired straight downward, its initial velocity is in the positive y-direction. At the moment it hits the ground, its final displacement (h) will be equal to the distance from the top of the cliff to the ground.
Let's calculate the time it takes for pellet B to hit the ground:
0 = 27.3t + (1/2)(-9.8)t²
0 = 27.3t - 4.9t²
To find the time, we solve this quadratic equation. Rearranging it to standard form:
4.9t² - 27.3t = 0
Factoring out common terms:
t(4.9t - 27.3) = 0
t = 0 or t = 27.3/4.9
Since time cannot be negative, we discard t = 0.
Thus, t = 27.3/4.9 ≈ 5.57 seconds is the time it takes for pellet B to hit the ground.
Now, we need to find out how long it takes for pellet A to hit the ground. Pellet A follows a projectile motion, going up first and then falling back down. The time it takes to reach the maximum height is the same as the time it takes to fall back down to the ground.
Since the initial velocity is 27.3 m/s and the acceleration is -9.8 m/s² (due to gravity, but in the opposite direction), we can use the equation:
v = u + at
Where:
v = final velocity (in this case, when the pellet reaches the maximum height, its velocity will be 0 m/s)
u = initial velocity
a = acceleration
t = time
0 = 27.3 - 9.8t
Solving for t:
9.8t = 27.3
t = 27.3/9.8 ≈ 2.79 seconds
So, it takes approximately 2.79 seconds for pellet A to reach the maximum height, and since it takes the same amount of time to fall back down to the ground, pellet A will hit the ground about 2.79 seconds after pellet B.