a projectile is shot vertically from ground level its height h in feet after t seconds is given by h=-16t^2+80t.How high will the projectile go?

Question

Please help quick I need this please gratefully appreciated

Answer 1

the projectile will reach its peak at the vertex of this quadratic function. The x coordinate of the vertex is at -b/2a=-80/-32= 5/2. now plug 5/2 into the original equation to find h

h = -16(5/2)sq + 80(2.5)

h = -16(25/4) + 200

h= 100

Answer 2

a projectile is shot vertically from ground level its height h in feet after t seconds is given by h=-16t^2+80t.How high will the projectile go?

The velocity of the the projectile is expressed by finding the derivative of the height wrt to time, or the change in height with time V = dh/dt , where V is the velocity

V = dh/dt = -32 t + 80 , now at maximum height the velocity is zero, then by substituting zero for velocity the time is: 0 = -32t + 80, solving for t , -32 t = -80, t = 80/32 = 2.5 sec

substituting t into h = -16 t² + 80 t = -16 ( 2.5)² + 80 ( 2.5) = 100 ft