A falling stone is at a certain instant 90 feet above the ground. Two seconds later it is only 10 feet above the ground. If it was thrown down with an initial speed of 5 feet per second, from what height was it thrown?

the initial height is 90.556 m

Step-by-step explanation:

neglecting air friction and the height required to accelerate the rock from 0 to 5 ft/s downwards. Then the only acceleration we are counting is the gravity and from the equations of vertical motion we have:

h₁=H+vy*t - 1/2*g*t₁²

h₂=H+vy*t - 1/2*g*t₂²

subtracting the first equation to the second one

h₂-h₁ = vy * (t₂-t₁) - 1/2*g * (t₂²-t₁²)

h₂-h₁ = vy * (t₂-t₁) - 1/2*g * (t₂-t₁) (t₂+t₁)

denoting Δt = t₂-t₁ and Δh=h₂-h₁

Δh=vy*Δt-1/2*g*Δt * (t₂+t₁)

Δh=vy*Δt - 1/2*g*Δt * (2*t₁+Δt)

1/2*g*Δt * (2*t₁+Δt) = vy*Δt-Δh

2*t₁+Δt = (vy*Δt-Δh) / (1/2*g*Δt)

t₁ = (vy*Δt-Δh) / (g*Δt) - Δt/2

replacing values

t₁ = (-5ft/s*2 s - (10 ft-90 ft)) / (32.2 ft/s²*2 s) - 2 s/2 = 0.087 s

then replacing in the first equation

h₁=H+vy*t - 1/2*g*t₁²

H = h₁ - vy*t + 1/2*g*t₁²

H = 90 ft - (-5ft/s) * 0.087 s + 1/2*32.2 ft/s² * (0.087 s) ² = 90.556 m

then the initial height is 90.556 m