Answer:
[tex]3.58\:\mathrm{s}[/tex]
Explanation:
We can use the kinematics equation [tex]\Delta y=v_it+\frac{1}{2}at^2[/tex] to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:
[tex]\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}[/tex]
Now we can substitute values in our kinematics equation:
- [tex]\Delta y=-100[/tex]
- [tex]a=-9.8\:\mathrm{m/s^2}[/tex] (acceleration due to gravity)
- [tex]v_i=-10.3283743813\:\mathrm{m/s}[/tex]
- Solving for [tex]t[/tex]
[tex]-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}[/tex]
