Question

a ball is thrown straight up at an initial speed of 7 meters per second. the height of the ball in meters follows the equation $h = -5t^2 + vt$, where $v$ is the initial speed and $h$ is the height at the given time. when will the ball first reach a height of 2m? equation: $-5t^2 + 7t - 2 = 0$ solution:?

Explanation:

Step1: Identify the quadratic equation

We have the quadratic equation \(-5t^{2}+7t - 2=0\). For a quadratic equation of the form \(ax^{2}+bx + c = 0\) (here \(x = t\), \(a=- 5\), \(b = 7\), \(c=-2\)), we can use the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) or factor the equation. Let's try factoring first. Multiply both sides by - 1 to get \(5t^{2}-7t + 2=0\). Now we need two numbers that multiply to \(5\times2 = 10\) and add up to - 7 (but since the middle term is - 7t and the constant term is + 2, we can rewrite it as \(5t^{2}-5t-2t + 2=0\)).

Step2: Factor the quadratic equation

Factor by grouping:
\(5t^{2}-5t-2t + 2=0\)
\(5t(t - 1)-2(t - 1)=0\)
\((5t - 2)(t - 1)=0\)

Step3: Solve for t

Set each factor equal to zero:
\(5t-2 = 0\) or \(t - 1=0\)
For \(5t-2=0\), we get \(t=\frac{2}{5}=0.4\)
For \(t - 1=0\), we get \(t = 1\)
Since we want the first time the ball reaches 2m, we take the smaller value of \(t\).

Answer:

The ball first reaches a height of 2m at \(t = 0.4\) seconds (or \(\frac{2}{5}\) seconds).