Math 110 College Algebra

Question

A projectile is launched straight up from ground level with an initial velocity of v0 per sec. Neglecting air resistance, its height in feet, s, after t seconds after launch is given by the equation s = –16t 2 + v0t.

(a) At what time(s) will the projectile reach a height of 112 feet when v0 = 96 feet/sec?

(b) After how many seconds will it return to the ground when v0 = 96 feet/sec?

Answer 1

As specified, the height is given by s = -16t² + V_ot

(a) with V_o = 96 feet/cm₂ amd s = 112 feet, putting these values in the above equation, we have

112 = -16t² + 96t

Solving for t in the above quadratic equation we get,

t² - 6t + 7 = 0.

t = 1.6 s and 4.4 s.

So the projectile will be ata a height of 112 feet at two times, at 1.6s and 4.4 secs.

(b) Differentiating the above height equation, we have

ds/dt = -32t + V_o

SO putting speed as 0, which means at the top of the projectile motion, and V_o = 96, we get t = 3 secs.

SO the projectile reaches the top at t = 3 secs. It will complete the motion at t = 6 secs.

So it will reach the ground at t = 6 secs.