A pellet gun is fired straight downward from the edge of a cliff that is 23.1 m above the ground. The pellet strikes the ground with a speed of 38.6 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

I get 99.118 but the answer is coming up wrong

To find the height the pellet would have reached if the gun was fired straight upward, we can use the conservation of energy.

The initial potential energy of the pellet at the cliff edge is given by mgh, where m is the mass of the pellet, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height of the cliff (23.1 m).

The initial kinetic energy of the pellet when it strikes the ground is given by (1/2)mv^2, where v is the speed of the pellet (38.6 m/s).

Since energy is conserved, the initial potential energy is equal to the final kinetic energy:

mgh = (1/2)mv^2

We can cancel out the mass (m) from both sides of the equation:

gh = (1/2)v^2

Now we can solve for h, which is the height the pellet would have reached if fired straight upward:

h = (1/2)(v^2)/g

Substituting the given values:

h = (1/2)(38.6^2)/9.8
h = 604.706 m

Therefore, the pellet would have gone approximately 604.706 m above the cliff edge if the gun was fired straight upward.

To find how far above the cliff edge the pellet would have gone if fired straight upward, we can use the concept of projectile motion.

When a projectile is fired upward, it will follow a parabolic path and reach its maximum height before falling back down. At its maximum height, the vertical component of its velocity becomes zero.

First, we need to find the time it takes for the pellet to reach the ground when fired downward.

Using the kinematic equation:
v^2 = u^2 + 2as

Where:
v = final velocity (38.6 m/s, but reversed in direction since it is going downward)
u = initial velocity (0 m/s since it is at rest initially)
a = acceleration due to gravity (-9.8 m/s^2, since it is acting downward)
s = total displacement (unknown, we want to find this)

Plugging in the values:
(0 m/s)^2 = (38.6 m/s)^2 + 2(-9.8 m/s^2)s
0 m^2/s^2 = 1493.96 m^2/s^2 - 19.6 m/s^2 s

Rearranging the equation to solve for s:
19.6 m/s^2 s = 1493.96 m^2/s^2
s = 1493.96 m^2/s^2 / 19.6 m/s^2
s = 76.18 m

So, when fired downward, the pellet takes approximately 76.18 meters to reach the ground.

Now, to find how far above the cliff edge the pellet would have gone if fired straight upward, we can assume that the upward and downward paths are symmetrical. This means that the maximum height reached by the pellet would be the same as the distance it took to reach the ground when fired downward.

Therefore, the pellet would have gone 76.18 meters above the cliff edge if fired straight upward.