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Aleks04 [339]
3 years ago
9

A stone that is dropped freely from rest traveled half its total height in the last second. With what velocity will it strike th

e ground?​
Physics
2 answers:
Paha777 [63]3 years ago
8 0

Answer:

33.5 m/s

Explanation:

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

h = 0 + ½ g (t + 1)²

h = ½ g (t + 1)²

Set equal and solve for t:

gt² = ½ g (t + 1)²

2t² = (t + 1)²

2t² = t² + 2t + 1

t² − 2t = 1

t² − 2t + 1 = 2

(t − 1)² = 2

t − 1 = ±√2

t = 1 ± √2

Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

v = at + v₀

v = g (2 + √2) + 0

v = g (2 + √2)

If g = 9.8 m/s², v = 33.5 m/s.

VashaNatasha [74]3 years ago
6 0
Let the total distance from the ground be “h” , so the distance travelled in last one second will be “ h/2”. So as it is free fall , u(initial velocity) = 0.
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