a ball is thrown straight up at an initial speed of 7 meters per second. the height of the ball in meters…

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:?

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:?

Answer

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.