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.

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.

lasttccasey · Answer

Assuming it was thrown straight down, the equation becomes:$$192 = 16t+16t^2$$From here solve for t using quadratic formula. Take the positive answer and you'll find that:$$t = 3s$$