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=32

Accepted Answer

Start with the given height function for the projectile: $$s = -16t^2$ + v_0$ t$$, where $$v_0 = 32$ feet per second. Substitute v_0$ into the equation to get s$$ = $$-16t^2 + 32t$. For part (a), set the height s$ equal to 80 feet to find the time(s) when the projectile reaches this height: -16t^2$ + 32t = 80. $$Rearrange the equation to standard quadratic form: $$-16t^2 + 32t - 80 = 0$. To simplify, divide the entire equation by -16, resulting in t^2$ - 2t + 5 = 0. $$Use the quadratic formula to solve for $$t$$: $$t = \frac{-b \pm \sqrt{b^2$ - 4ac}}{2a}$$, where $$a=1$$, $$b=-2$$, and $$c=5. $$Calculate the discriminant $$b^2 - 4ac$ to determine the nature of the roots. For part (b), find the time when the projectile returns to the ground by setting s=0$: -16t^2$ + 32t = 0. $$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