31 velocity

Question

You throw a ball in the air with an initial velocity of 31 feet per second. The ball leaves your hand when it is 6 feet above the ground. You catch the ball when it reaches a height of 4 feet. After how many seconds do you catch the ball.

Answer 1

You can use an equation which represents the height of any object that travels up and down (as a projectile) over the course of time, given the initial velocity and the height at which it started. That equation is:

 

(height at the end of a specific time) = (-16)(time)² + (initial velocity)(time) + (initial height)

 

for this situation, that would be:

 

(4) = (-16)(time)² + (31)(time) + 6

 

What you'd want to do is to subtract 4 from both sides of the equation to get:

 

0 = -16T² + 31T + 2

 

Then you can solve this by factoring:

 

0 = (-16T - 1) (T - 2)

 

so T = -1/16   and also T = 2    (the negative answer isn't possible in this situation, so the answer is "in 2 seconds")