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A ball is thrown straight down from the top of a building at a velocity of 16 ft/s. The building is 480 feet tall, and the acceleration due to gravity is 32 ft/s2.

This problem can be represented using the following equation.

How much time will the ball take to reach the ground? (1/2)32t^2+16t=480

User EOnOe
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1 Answer

9 votes
9 votes

Answer:

Time, t = 5 seconds.

Explanation:

Given the following data;

Initial velocity = 16 ft/s

Height = 480 ft

Acceleration due to gravity = 32 ft/s²

To find how much time it will take the ball to reach the ground, we would use the kinematic equation given by the formula;


d = V_(i)t + \frac {1}{2}at^(2)

Where;

  • d is the displacement.
  • Vi is the initial velocity.
  • a is the acceleration or acceleration due to gravity.
  • t is the time measured in seconds.

Substituting into the equation, we have;


480 = 16t + \frac {1}{2}*32t^(2)


480 = 16t + 16t^(2)

Dividing all through by 16, we have;


30 = t + t^(2)

Re-arranging the quadratic equation, we have;


t^(2) + t - 30 = 0

We would solve the quadratic equation by factorization;


t^(2) + 6t - 5t - 30 = 0


t(t + 6) - 5(t + 6) = 0


(t + 6)(t - 5) = 0

t = -6 or 5

Since we know time can't be negative, we would ignore the value of -6.

Therefore, time = 5 seconds

User Michal Kottman
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3.0k points