Answer:
9.6 seconds
Step-by-step explanation:
The equation for projectile motion in feet is [tex]h(t)=-16t^2+v_0t+h_0[/tex] where [tex]v_0[/tex] is the initial velocity in ft/s and [tex]h_0[/tex] is the initial height in feet.
We are given that the initial upward velocity is [tex]v_0=150ft/sec[/tex] and the initial height is [tex]h_0=35ft[/tex]. Thus, plugging these values into our equation, we get [tex]h(t)=-16t^2+150t+35[/tex] as our equation for the situation.
The firework will land, assuming it doesn't explode, when [tex]h(t)=0[/tex], thus:
[tex]h(t)=-16t^2+150t+35\\\\0=-16t^2+150t+35\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-150\pm\sqrt{150^2-4(-16)(35)}}{2(-16)}\\ \\t=\frac{-150\pm\sqrt{22500+2240}}{-32}\\\\t=\frac{-150\pm\sqrt{24740}}{-32}\\\\t=\frac{-150\pm2\sqrt{6185}}{-32}\\ \\t=\frac{75\pm\sqrt{6185}}{16}\\\\t_1\approx-0.2\\\\t_2\approx9.6[/tex]
Since time can't be negative, then the firework will land after 9.6 seconds, assuming it doesn't explode at the time of landing.