A pellet gun is fired straight downward from the edge of a cliff that is 27.1 m above the ground. The pellet strikes the ground with a speed of 25.9 m/s. How far above the cliff edge would the pellet have gone had the gun been fired straight upward?

vf^2=vi^2+2gd where g=-9.8, d=-27.1

solve for vi.

Now it it had been straight up,
0=Vi^2+2gh solve for h. g=-9.8

To determine how far above the cliff edge the pellet would have gone if the gun was fired straight upward, we need to use the principle of conservation of energy.

When the pellet is fired downward, it loses gravitational potential energy and gains kinetic energy. The initial potential energy of the pellet is given by the equation:

PE_initial = m * g * h

Where:
m is the mass of the pellet
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height of the cliff (27.1 m)

The initial kinetic energy of the pellet can be calculated using the equation:

KE_initial = 0.5 * m * v^2

Where:
v is the velocity of the pellet (25.9 m/s)

Since energy is conserved, the initial potential energy of the pellet is converted entirely into kinetic energy when it strikes the ground. Therefore, the final kinetic energy can be equated to the initial potential energy:

KE_final = PE_initial

Substituting the equations for kinetic and potential energy:

0.5 * m * v^2 = m * g * h

Canceling out the mass of the pellet:

0.5 * v^2 = g * h

Now, we can rearrange the equation to find the height the pellet would reach when fired straight upward:

h_upward = 0.5 * v^2 / g

Substituting the values given:

h_upward = 0.5 * (25.9 m/s)^2 / 9.8 m/s^2

Calculating the result:

h_upward = 3.4245... m

Therefore, the pellet would have gone approximately 3.42 meters above the cliff edge if fired straight upward.