Question

Height of a Projectile A projectile is launched from ground level with an initial velocity of v0 feet per second. Neglecting air resistance, its height in feet t seconds after launch is given by s=-16t2+v0t. In each exercise, find the time(s) that the projectile will (a) reach a height of 80 ft and (b) return to the ground for the given value of v0. Round answers to the nearest hundredth if necessary. v0=96

Accepted Answer

Start with the given height function for the projectile: $$s = -16t^2$ + v_0$ t$$, where $$v_0 = 96$ feet per second. Substitute v_0$ into the equation to get s$$ = $$-16t^2 + 96t$. For part (a), set the height s$ equal to 80 feet to find the time(s) when the projectile reaches this height: -16t^2$ + 96t = 80. $$Rearrange the equation to standard quadratic form: $$-16t^2 + 96t - 80 = 0$. You can simplify this equation by dividing all terms by -16 to make calculations easier. Solve the quadratic equation for t$ using the quadratic formula: t$$ = \frac{-b \pm \sqrt{$$b^2 - 4ac$$}}{2a}$$, where a$, b$, and c$ are the coefficients from the quadratic equation. For part (b), find the time when the projectile returns to the ground by setting s = 0$ and solving -16t^2$ + 96t = 0. $$Factor the equation and solve for $$t to $$find the time(s) when the height is zero.

Verified video answer for a similar problem:

Key Concepts

Quadratic Functions and Their Graphs

Solving Quadratic Equations

Interpreting Solutions in Context