Question

How the heck do I solve this? Please help T^T

A basketball is thrown upward from a height of 1.8 m at a velocity of 12 m/s. How long until it reaches the ground?

(Recall: h(t)=-4.9t^2-v_0 t+h_0, where v_0 is initial velocity and h_0 is initial height.)

Answer 1

Explanation:

The quadratic is

`s(t)=-4.9t^2+12t+1.8`

where s(t) is the height of the ball after a certain amount of time goes by. If we are looking for how long til it hits the ground, we know 2 things from that question. First we know that the height of an object on the ground is 0, and we also know that t is the time that the ball is in the air. Because the height of the ball when it's on the ground is 0, we will set s(t) equal to 0 and factor the quadratic to solve for t, the time the ball is in the air. Throw it into the quadratic formula to find those times.

`t=(-12+/-√(144-4(-4.9)(1.8)) )/(2(-4.9))` which will simplify to

`t=(-12+/-√(179.28) )/(-9.8)`

The 2 solutions we get from that are that

`t=(-12+√(179.28) )/(-9.8)=-.1417906 sec` and

`t=(-12-√(179.28) )/(-9.8)=2.59077sec`

Now, the 2 things in math that will never be negative are distances and time, so we will disregard the negative time and go with t = 2.59 seconds. Not sure to where you need to round.