Apellet gun is fired straight downward from the edge of a cliff that is 12.7 m above the ground. the pellet strikes the ground with a speed of 27.9 m/s. how far above the cliff edge would the pellet have gone had the gun been fired straight upward?

h₂ = 27.22m : height the pellet reaches

Explanation:

Pellet kinetics

Pellet moves with uniformly accelerated movement in  

freefall  

v f₁²=v₀²+2g*h₁ (formula 1)  : ball movement down

v f₂²=v₀²-2g*h₂ (formula (formula 2)  : ball movement up

d:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

g: acceleration due to gravity in m/s²

h :height in m

Known data  

g=9.8 m/s²

h₁=12.7m

v f₁= 27.9 m/s

Initial velocity calculation (v₀)

We replace the data in formula 1:

27.9²=v₀²+2*9.8* 12.7

27.9²-2*9.8* 12.7=v₀²

529.49=v₀²

Calculation of the height that the pellet reaches when it goes up

Data:

v f₂=0 :When the pellet reaches the maximum height the final speed is zero

We replace the data in formula (2)

0²=23.1²-2*9.8*h₂

19.6h₂ =23.1²

h₂ =23.1²÷19.6

h₂ =27.22m

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered