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 equals negative 16 t squared plus v 0 ts=−16t2+v0t. Find the​ time(s) that the projectile will​ (a) reach a height of 192 ft and​ (b) return to the ground when v 0v0equals=112 feet per second.

Answer 1

U just need to put value of h=192 feet in equation
h==−16t2+v0t
192=-16t^2
here v0=0

now find h

b) when objects reach to ground then h=0
so same equation put v0=112 feet

Answer 2

Step-by-step explanation:

It is given that, a projectile is launched from ground level. Its height as a function of time is given by :

$h(t)=-16t^2+v_ot$...............(1)

Where

v₀ is the initial velocity of the projectile, $v_o=112\ ft/s$

(a) Let t is the time taken by the projectile to reach a height of 192 ft. Equation (1) becomes :

$-16t^2+112t=192$

Aftyer solving the above equation we get :

t = 3 seconds (on the way up)

and t = 4 seconds (on the way down)

(b) When the projectile returns to the ground, h(t) = 0                  

$-16t^2+112t=0$  

On solving the above quadratic equation, t = 7 seconds

Hence, this is the required solution.