Question

An object falls a distance h from rest. If it travels 0.440h in the last 1.00 s, find (a) the time and (b) the height of its fall.

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Initial speed of the object = V₀ = 0 m/s (At rest)

Initial height of the object = H

Time taken by the object to reach the ground = T

H = V₀T + gT²/2

H = (0)T + gT²/2

H = gT²/2

Speed of the object 1 sec before hitting the ground = V₁

V₁ = V₀ + g(T - 1)

V₁ = 0 + g(T - 1)

V₁ = gT - g

Speed of the object just before hitting the ground = V₂

V₂ = V₀ + gT

V₂ = 0 + gT

V₂ = gT

Distance through which the object falls in the last second = L = 0.44H = 0.44(gT²/2) = 0.22gT²

V₂² = V₁² + 2gL

(gT)² = (gT - g)² + 2g(0.22gT²)

g²T² = g²T² - 2g²T + g² + 0.44g²T²

T² = T² - 2T + 1 + 0.44T²

0.44T² - 2T + 1 = 0

T = 3.97 sec or 0.57 sec

Time period cannot be lower than 1 sec.

T = 3.97 sec

H = gT²/2

H = (9.81)(3.97)²/2

H = 77.3 m

a) Time of the fall = 3.97 sec

b) Height of the fall = 77.3 m