Question

Make an equation and solve x? (Quadratic Equations)

a stone is thrown vertically upwards from the ground. The height of the stone reaches in `t` seconds, given by `y=30t-5t^2`.

a) How long does it take for the stone to reach the ground again?

b) How long does it take for the stone to reach a height of 25 metres?

Answer 1

a) the stone reaches the ground again at `t=6`
b) the stone reaches `y=25` at `t=1`

Explanation:

First, we assume that the ground is at `y=0`, so part a) is asking when this happens after the initial throw. We could solve this using the quadratic formula, but this time is simple enough for us to solve it by factoring. Let's re-write the equation by factoring a `t` out on the right hand side:

`y=t*(30-5t)`

This shows us that there are two solutions to `y=0`, first when `t=0` (which is the initial throw) and next when:

`30-5t = 0 implies t=6`

Part b) asks us to solve for `t` when `y=25`:

`25 = 30t-5t^2`

This time we will use the quadratic formula so we need to put the equation into standard form:

`0=-5t^2+30t-25`

`t = (-30 +- sqrt(30^2-4(-5)(-25)))/(2( -5))`

`t = 3 +- 2`

`t = 1, 5`

Graphing the equation we see that the curve does cross `y=25` twice, once on the way up at `t=1` and then on the way down at `t=5`

graph{30x-5x^2 [-1, 7, -3, 50]}