t = 5.05 sec
Solution:
Given data:
Initial height [tex]\left(h_{0}\right)[/tex] = 24 meter
Initial velocity [tex]\left(v_{0}\right)[/tex] = 22 meter
Height of the rock (h) = 10 meter
To find when will the height of the rock be 10 metres:
General formula for the height(meter) of a projectile over time:
[tex]h(t)=-4.9 t^{2}+v_{0} t+h_{0}[/tex]
Substitute the given data in the above formula.
[tex]10=-4.9 t^{2}+22 t+24[/tex]
[tex]-4.9 t^{2}+22 t+24-10=0[/tex]
[tex]-4.9 t^{2}+22 t+14=0[/tex]
This equation looks like a quadratic equation.
We can solve it by using the quadratic formula:
[tex]$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}[/tex]
Here a = –4.9, b = 22, c = 14
[tex]\begin{aligned}t &=\frac{-22 \pm \sqrt{(22)^{2}-4(-4.9)(14)}}{2(-4.9)} \\&=\frac{-22 \pm \sqrt{484+274.4}}{-9.8} \\&=\frac{-22 \pm \sqrt{758.4}}{-9.8} \\&=\frac{-22 \pm {27.54}}{-9.8} \\t &=\frac{-22-27.54}{-9.8}\ (\text {or})\ t=\frac{-22+27.54}{-9.8}\\t &=5.05\ (\text {or})\ t=-0.56\end{aligned}[/tex]
Time cannot be indicated in negative. So neglect t = –0.56.
Therefore, t = 5.05 sec
Hence after 5.05 sec the rock will be 10 meters from ground level.
