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12 February, 20:14

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

+1
Answers (1)
  1. 12 February, 23:00
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    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)
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