Question

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. Find the time(s) that the projectile will (a) reach a height of 288 ft and (b) return to the ground when v0=144 feet per second.

Answer 1

Explanation:

I've posted solutions in the picture. Rather the ways to solve them. Check and find answers on your own.

Also, I've not solved the problem by differential calculus. You can, obviously, if you're interested. Use it for ease and for tougher equations.

Answer 2

The solution is:

a) time to reach s = 288 ft is 3 sec

b) Total time of the fly is 9 sec

a) The height s = - 16×t² + v₀×t

At a height of 288 ft with v₀ = 144 ft/sec

288 = -16t² + 144×t or 16×t² - 144×t + 288 = 0

Simplifying dividing by 16 t² - 9×t + 18 = 0

We got a second degree equation, solving for t

t₁,₂ = [ 9 ±√81 - 72 ]/2

t₁ = (9 + 3)/2 t₁ = 6 sec and t₂ = ( 9 - 3 ) /2 t₂ = 3 sec

we got 2 possible solution, we will examine s maximum

when s is maximum ds/dt = 0 then

s = - 16×t² + v₀×t tacking derivatives on both sides of the equation

ds/dt = - 32× t + v₀ ds/dt = 0 = - 32× t + 144

- 32×t + 144 = 0 t = 144/32 t = 4,5 sec

And s max = - 16 ×( 4,5)² + 144 ( 4,5)

s maximum is s = - 324 + 648 s max = 324 ft

Then if s maximum will occurs at 4,5 sec and it was 324 ft, it is obviuos that time to get 288 ft was with t₂ = 3 sec.

b) As the time to get s maximum is 4,5 sec, by symmetry total time is 9 sec