Question

Use the formula d= vt + 16t2, where d is the distance in feet, v is the initial velocity in feet per second, and it is the time in seconds.

An object is released from the top of a building 480 ft high. The initial velocity is 16 ft/s. How many seconds later will the object hit the ground?

Answer 1

• 5 seconds.

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As per given we have:

• d = 480 ft,
• v = 16 ft/s.

Substitute these into given equation and solve for t:

• 480 = 16t + 16t²
• 30 = t + t²
• t² + t - 30 = 0
• t² + 6t - 5t - 30 = 0
• t(t + 6) - 5(t + 6) = 0
• (t + 6)(t - 5) = 0
• t + 6 = 0 or t - 5 = 0
• t = - 6 or t = 5

The first value of t is discarded as time can't be negative, so the answer is 5 seconds.

Answer 2

5 seconds

Explanation:

d = vt + 16t² , that is

vt + 16t² = d

substitute d = 480, v = 16 into the equation

16t + 16t² = 480 ( divide through by 16 )

t + t² = 30 ( subtract 30 from both sides )

t² + t - 30 = 0 ← in standard form

(t + 6)(t - 5) = 0 ← in factored form

equate each factor to zero and solve for t

t + 6 = 0 ⇒ t = - 6

t - 5 = 0 ⇒ t = 5

t > 0 , then t = 5

the object will hit the ground 5 seconds later