while warming up at soccer practice, ellies teammate tosses the ball to ellie. ellie hits the ball upward…

while warming up at soccer practice, ellies teammate tosses the ball to ellie. ellie hits the ball upward with her head from a height of 1.6 meters at a velocity of 6 meters per second. the ball is in the air for some time until ellie catches it at a height of 1 meter. to the nearest tenth of a second, how long is the ball in the air before ellie catches it? hint: use the formula h = -4.9t² + vt + s.

while warming up at soccer practice, ellies teammate tosses the ball to ellie. ellie hits the ball upward with her head from a height of 1.6 meters at a velocity of 6 meters per second. the ball is in the air for some time until ellie catches it at a height of 1 meter. to the nearest tenth of a second, how long is the ball in the air before ellie catches it? hint: use the formula h = -4.9t² + vt + s.

Answer

Explanation:

Step1: Identify values

Given $h = 1$, $v = 6$, $s=1.6$, substitute into $h=-4.9t^{2}+vt + s$. So we get $1=-4.9t^{2}+6t + 1.6$.

Step2: Rearrange to quadratic form

Rearrange the equation to $4.9t^{2}-6t - 0.6=0$. Here $a = 4.9$, $b=-6$, $c=-0.6$.

Step3: Use quadratic formula

The quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a$, $b$, $c$ values: $t=\frac{6\pm\sqrt{(-6)^{2}-4\times4.9\times(-0.6)}}{2\times4.9}=\frac{6\pm\sqrt{36 + 11.76}}{9.8}=\frac{6\pm\sqrt{47.76}}{9.8}$.

Step4: Calculate values of t

$\sqrt{47.76}\approx6.91$. Then $t=\frac{6\pm6.91}{9.8}$. We have two solutions for $t$: $t_1=\frac{6 + 6.91}{9.8}\approx1.3$ and $t_2=\frac{6-6.91}{9.8}\approx - 0.1$. Since time cannot be negative, we discard $t_2$.

Answer:

$1.3$