The following questions present a twist on the scenario above to test your understanding.

Suppose another stone is thrown horizontally from the same building. If it strikes the ground 67 m away, find the following values.

(a) time of flight

s

(b) initial speed

m/s

(c) speed and angle with respect to the horizontal of the velocity vector at impact

m/s

a) t = √ (2*h/g)

b) v₀ = 67*√ (g / (2*h))

c) ∅ = tan⁻¹ (2*h/67)

Explanation:

Suppose that the height of the building is known (h), then we have

a) y = h = g*t²/2

then the time of flight is

t = √ (2*h/g)

b) The initial speed (v₀) can be obtained as follows

x = v₀*t ⇒ v₀ = x / t

where

x = xmax = 67 m and t = √ (2*h/g)

So, we have

v₀ = 67 / √ (2*h/g) = 67*√ (g / (2*h))

c) In order to get the speed and the angle with respect to the horizontal of the velocity vector at impact, we obtain vx and vy as follows

vx = v₀ = 67*√ (g / (2*h))

and

vy = gt = g*√ (2*h/g) ⇒ vy = √ (2*h*g)

then we apply

v = √ (vx² + vy²) = √ ((67*√ (g / (2*h))) ² + (√ (2*h*g)) ²)

⇒ v = √ (g * (4489 + 4*h²) / (2*h))

The angle is obtained as follows

tan ∅ = vy / vx ⇒ tan ∅ = √ (2*h*g) / 67*√ (g / (2*h))

⇒ tan ∅ = 2*h / 67 ⇒ ∅ = tan⁻¹ (2*h/67)