Question

Let us suppose that one of the fireworks is launched from the top of the building with an initial upward velocity of 150 ft/s and the building has a height of 35 feet. what is the equation for this situation?

Answer 1

Answer:

9.6 seconds

Step-by-step explanation:

The equation for projectile motion in feet is $h(t)=-16t^2+v_0t+h_0$ where $v_0$ is the initial velocity in ft/s and $h_0$ is the initial height in feet.

We are given that the initial upward velocity is $v_0=150ft/sec$ and the initial height is $h_0=35ft$. Thus, plugging these values into our equation, we get $h(t)=-16t^2+150t+35$ as our equation for the situation.

The firework will land, assuming it doesn't explode, when $h(t)=0$, thus:

$h(t)=-16t^2+150t+35\\\\0=-16t^2+150t+35\\\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \\t=\frac{-150\pm\sqrt{150^2-4(-16)(35)}}{2(-16)}\\ \\t=\frac{-150\pm\sqrt{22500+2240}}{-32}\\\\t=\frac{-150\pm\sqrt{24740}}{-32}\\\\t=\frac{-150\pm2\sqrt{6185}}{-32}\\ \\t=\frac{75\pm\sqrt{6185}}{16}\\\\t_1\approx-0.2\\\\t_2\approx9.6$

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.