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 \(-5t^{2}+7t - 2=0\). Multiply both sides by \(- 1\) to make the coefficient of \(t^{2}\) positive: \(5t^{2}-7t + 2=0\)

Step2: Use the quadratic formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the quadratic formula is \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 5\), \(b=-7\), \(c = 2\)

First, calculate the discriminant \(\Delta=b^{2}-4ac=(-7)^{2}-4\times5\times2=49 - 40 = 9\)

Then, \(t=\frac{7\pm\sqrt{9}}{2\times5}=\frac{7\pm3}{10}\)

Step3: Find the two solutions

We have two solutions:
\(t_{1}=\frac{7 + 3}{10}=\frac{10}{10}=1\)
\(t_{2}=\frac{7-3}{10}=\frac{4}{10}=0.4\)

Since we want the first time the ball reaches \(2\)m, we take the smaller value of \(t\)

Answer:

The ball first reaches a height of \(2\)m at \(t = 0.4\) seconds.