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?

Explanation:

Step1: Substitute values into the equation

We know \( h = 2 \), \( v = 7 \), and the equation \( h=-5t^{2}+vt \). Substitute these values:
\( 2=-5t^{2}+7t \)
Rearrange to standard quadratic form \( ax^{2}+bx + c = 0 \):
\( 5t^{2}-7t + 2 = 0 \)

Step2: Solve the quadratic equation

Use the quadratic formula \( t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 5 \), \( b=-7 \), \( c = 2 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-7)^{2}-4\times5\times2=49 - 40 = 9 \)
Then, \( t=\frac{7\pm\sqrt{9}}{10}=\frac{7\pm3}{10} \)
We get two solutions: \( t_1=\frac{7 + 3}{10}=1 \) and \( t_2=\frac{7-3}{10}=0.4 \)

Step3: Determine the first time

Since we want the first time the ball reaches 2m, we choose the smaller positive solution.

Answer:

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