A pellet gun is fired straight downward from the edge of a cliff that is 13.4 m above the ground. The pellet strikes the ground with a speed of 26.1 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?
v = v0 - 9.8t
v0 = -26.1-9.8t
h = 13.4 + v0*t - 4.9t^2
= 13.4 + (-26.1-9.8t)t - 4.9t^2
h=0 when t=0.416
v0 = -26.1-9.8*0.416= -30.17
So, if v0 = +30.17,
h = 13.4 + 30.17t - 4.9t^2
max height at t = 30.17/9.8 = 3.078
h(3.078) = 59.8m
Note that that is not the answer to the question ...
To find the height above the cliff edge that the pellet would have reached if fired straight upward, we need to use the concept of projectile motion.
In projectile motion, the vertical motion and horizontal motion are independent of each other. Therefore, we can analyze the vertical motion of the pellet separately.
The initial velocity of the pellet when fired straight downward is 0 m/s, as it's dropped from rest. The acceleration due to gravity (g) is approximately 9.8 m/s², acting downwards.
We can use the following kinematic equation to find the time it takes for the pellet to reach the ground:
h = (1/2) * g * t²
Where:
h = height above the ground (13.4 m)
g = acceleration due to gravity (9.8 m/s²)
t = time
Rearranging the equation to solve for t:
t = sqrt((2 * h) / g)
Plugging in the given values:
t = sqrt((2 * 13.4) / 9.8) ≈ 1.86 seconds
Now, if the gun had been fired straight upward, the time it takes for the pellet to reach its maximum height would be the same as the time it takes to fall back down to the ground.
Therefore, the total time of flight for the pellet fired straight upward is approximately 2 * 1.86 seconds = 3.72 seconds.
Using this time of flight, we can find the maximum height above the ground reached by the pellet:
h_max = (1/2) * g * t_max²
Where:
h_max = maximum height above the ground (unknown)
g = acceleration due to gravity (9.8 m/s²)
t_max = time of flight (3.72 seconds)
Rearranging the equation to solve for h_max:
h_max = (1/2) * g * t_max²
Plugging in the given values:
h_max = (1/2) * 9.8 * (3.72)² ≈ 66.93 meters
Therefore, if the gun had been fired straight upward, the pellet would have gone approximately 66.93 meters above the cliff edge.