-45m=(10m/s)t-(1)/(2)(9,80m/s^2)t^2

To solve the equation \(-45m = (10 \, \text{m/s})t - \frac{1}{2}(9.80 \, \text{m/s}^2)t^2\), we need to find the value of \(t\) that satisfies the equation. This is a quadratic equation in the form of \(at^2 + bt + c = 0\), where:<br /><br />- \(a = -\frac{1}{2}(9.80 \, \text{m/s}^2) = -4.9 \, \text{m/s}^2\)<br />- \(b = 10 \, \text{m/s}\)<br />- \(c = 45 \, \text{m}\)<br /><br />We can solve this quadratic equation using the quadratic formula:<br /><br />\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]<br /><br />First, let's calculate the discriminant (\(\Delta\)):<br /><br />\[ \Delta = b^2 - 4ac \]<br />\[ \Delta = (10 \, \text{m/s})^2 - 4(-4.9 \, \text{m/s}^2)(45 \, \text{m}) \]<br />\[ \Delta = 100 \, \text{m}^2/\text{s}^2 + 882 \, \text{m}^2/\text{s}^2 \]<br />\[ \Delta = 982 \, \text{m}^2/\text{s}^2 \]<br /><br />Now, we can find the values of \(t\):<br /><br />\[ t = \frac{-10 \, \text{m/s} \pm \sqrt{982 \, \text{m}^2/\text{s}^2}}{2(-4.9 \, \text{m/s}^2)} \]<br />\[ t = \frac{-10 \, \text{m/s} \pm 31.37 \, \text{m/s}}{-9.8 \, \text{m/s}^2} \]<br /><br />This gives us two solutions for \(t\):<br /><br />\[ t_1 = \frac{-10 + 31.37}{-9.8} \]<br />\[ t_1 = \frac{21.37}{-9.8} \]<br />\[ t_1 \approx -2.18 \, \text{s} \]<br /><br />\[ t_2 = \frac{-10 - 31.37}{-9.8} \]<br />\[ t_2 = \frac{-41.37}{-9.8} \]<br />\[ t_2 \approx 4.23 \, \text{s} \]<br /><br />Since time cannot be negative in this context, we discard the negative solution. Therefore, the time \(t\) that satisfies the equation is approximately:<br /><br />\[ t \approx 4.23 \, \text{s} \]