Question

An object is thrown downward, with an initial speed of 16 ft/s, from the top of a building 192 ft high. How many seconds later will the object hit the ground? Use the equation

d = vt + 16t2,
where d is the distance in feet, v is the initial speed, and t is the time in seconds.

Answer 1

The object will hit the ground after 3 seconds.

Step-by-step explanation:

To determine the number of seconds it will take for the object to hit the ground, we can use the equation d = vt + 16t^2, where d is the distance, v is the initial speed, and t is time. In this case, the initial speed (v) is 16 ft/s and the distance (d) is 192 ft. Plugging in these values, we get 192 = 16t + 16t^2.

Rearranging the equation into quadratic form, we have 16t^2 + 16t - 192 = 0. To solve for t, we can use the quadratic formula or factor the equation. Factoring, we have (4t - 12)(4t + 16) = 0. So, t = 3 or t = -4. Taking the positive value, the object will hit the ground after 3 seconds.

Answer 2

If the speed is constant and never reaching terminal velocity then the object will hit the ground in 12 seconds.