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 = -5t² + 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?

Explanation:

Step1: Substitute values into the equation

We know \( h = 2 \), \( v = 7 \), and the equation \( h=-5t^{2}+vt \). Substituting these values in, we get:
\( 2=-5t^{2}+7t \)

Step2: Rearrange into standard quadratic form

Rearrange the equation to \( 5t^{2}-7t + 2 = 0 \) (by adding \( 5t^{2}-7t \) to both sides and then multiplying both sides by -1 to make the coefficient of \( t^{2} \) positive).

Step3: Solve the quadratic equation

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 \( D=b^{2}-4ac=(-7)^{2}-4\times5\times2=49 - 40 = 9 \)
Then, \( t=\frac{7\pm\sqrt{9}}{10}=\frac{7\pm3}{10} \)
We get two solutions: \( t_{1}=\frac{7 + 3}{10}=1 \) and \( t_{2}=\frac{7-3}{10}=\frac{4}{10}=0.4 \)
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.