# 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).

samhouston | Middle School Teacher | (Level 1) Associate Educator

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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