The height of a ball thrown straight up into the air as a function of time is given as h(t) = 1 + 15t Ð 4.9t^2. At what time does the ball hit the ground?

At ground level, h = 0

1+15t-4.9t^2 = 0
4.9t^2 - 15t - 1 = 0

Solve using the quadratic equation, using only the positive solution.

Thanks Reiny

To determine at what time the ball hits the ground, we need to find the value of t when the height of the ball, h(t), becomes 0.

We are given the equation for the height of the ball as a function of time: h(t) = 1 + 15t - 4.9t^2.

Setting h(t) equal to 0, we have:
0 = 1 + 15t - 4.9t^2.

Now we can solve this quadratic equation for t. One way to solve it is by factoring. However, in this case, let's solve it using the quadratic formula:

The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / 2a.

In our equation, a = -4.9, b = 15, and c = 1.

Plugging the values into the quadratic formula, we get:
t = (-15 ± √(15^2 - 4(-4.9)(1))) / (2(-4.9)).

Simplifying further:
t = (-15 ± √(225 + 19.6)) / (-9.8).

t = (-15 ± √(244.6)) / (-9.8).

Now, we have two possible values for t: one with positive square root and one with negative square root.

Calculating the positive square root:
t₁ = (-15 + √(244.6)) / (-9.8).

Calculating the negative square root:
t₂ = (-15 - √(244.6)) / (-9.8).

These two values represent two distinct times when the ball is at a height of 0. The positive value represents when the ball is on its way up, and the negative value represents when it's on its way down. We're interested in the time when the ball hits the ground, so we need to find the positive time, t₁.

Calculating the value of t₁ will give us the answer to when the ball hits the ground.