Question

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

Answer 1

Answer:

Time, t = 5 seconds.

Step-by-step 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