Respuesta :
The ball's height [tex]y[/tex] at time [tex]t[/tex] is given by
[tex]y = 29.2\,\mathrm m - \left(8.30\dfrac{\rm m}{\rm s}\right) t - \dfrac12 gt^2[/tex]
where [tex]g=9.80\frac{\rm m}{\mathrm s^2}[/tex].
Solve for [tex]t[/tex] when [tex]y=0[/tex]. Omitting the units, we have
[tex]29.2 - 8.30t - \dfrac g2 t^2 = 0[/tex]
I'll solve by completing the square.
[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t\right) = 0[/tex]
[tex]29.2 - \dfrac g2 \left(t^2 + \dfrac{16.6}g t + \dfrac{8.3^2}{g^2}\right) = -\dfracg2 \times \dfrac{8.3^2}{g^2}[/tex]
[tex]29.2 - \dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = -\dfrac{8.3^2}{2g}[/tex]
[tex]\dfrac g2 \left(t + \dfrac{8.3}g\right)^2 = 29.2 + \dfrac{8.3^2}{2g}[/tex]
[tex]\left(t + \dfrac{8.3}g\right)^2 = \dfrac{58.4}g + \dfrac{8.3^2}{g^2}[/tex]
[tex]t + \dfrac{8.3}g = \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]
[tex]t = -\dfrac{8.3}g \pm \sqrt{\dfrac{58.4}g + \dfrac{8.3^2}{g^2}}[/tex]
[tex]\implies t \approx -3.43 \text{ or } t \approx 1.74[/tex]
Ignore the negative solution; the ball hits the ground about 1.74 s after being thrown.
