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

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