Aperson standing on the edge of a high cliff throws a rock straight up with an initial velocity, v0v0, of 13.0m/s13.0m/s. on the way back down the rock misses the edge of the cliff and continues to fall. ignoring the effects of air resistance, calculate the position and velocity of the rock a

For the position, we can use the equation below for our calculations:
H = h + v₀²/2g
where 
H is the total height from the base of the cliff to the highest point of the rock on air
h is the height of the cliff
v₀ is the initial height
g is equal to 9.81 m/s²

H = h + 13²/2(9.81)
H = h + 8.61 m

Because the height of the cliff is not known, the answer would just be h +8.61 m.

For the velocity, we use the equation: v = √2gH. Substituting the answer for H, the velocity is

v = √2(9.81)(h+8.61)
v = 4.43√(h+8.61)

It may possibly be c i think |-/