# 1. If an object is thrown downward at 13 m/s from the top of a building, the equation relating displacement [metres] and time [seconds] is s= 13t+1/2 gt^2 . Find the time for an...

1. If an object is thrown downward at 13 m/s from the top of a building, the equation relating displacement [metres] and time [seconds] is s= 13t+1/2 gt^2 . Find the time for an object to reach the ground from the top of a 62.5 m high building (s = 62.5 m, g = 9.81 m/s2).

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### 1 Answer

s = 13 * t + 0.5 * g * t^2

(s = 62.5 m, g = 9.81 m/s^2)

**Step 1: Substitute**

62.5 = 13 * t + 0.5 * 9.81 * t^2

62.5 = 13 * t + 4.905 * t^2

**Step 2: Write quadratic equation in standard form**

4.905t^2 + 13t + -62.5 = 0

**Step 3: Identify "a", "b", and "c" for the quadratic formula**

a = 4.905, b = 13, c = -62.5

**Step 4: Apply the quadratic formula**

t = [-b `+-` √ (b^2 - 4ac)] / 2a

This is a complicated formula, so break it down into parts.

√ (b^2 - 4ac)

√ (13^2 - 4 * 4.905 * -62.5)

√ (169 - (-1226.25))

√ (1395.25) ≈ 37.353

The numerator is now -13 ± 37.353

-13 + 37.353 = 24.353 -13 – 37.353 = -50.353

The denominator is 2a

2 * 4.905 = 9.81

Now the two fractions are:

24.353 / 9.81 -50.353 / 9.81

≈ 2.482 ≈ -5.133

Since -5.133 seconds does not make sense as the time for an object to reach the ground, the solution is **≈ 2.482 seconds**.

Graphic Proof:

This is the graph of the equation:

t = 4.905x^2 + 13x + -62.5

Noticed the parabola crosses the x-axis at **≈ 2.482 seconds**