A projectile is fired at an inclination of 45 to the horizon

A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 96 feet per second. The height h of the projectile is given by the following where x is the horizontal distance of the projectile from the firing point. h(x) = (-32x^2)/(96^2) + x (a) How far from the firing point is the height of the projectile a maximum? Give your answer correct to the nearest foot. (b) Find the maximum height of the projectile. Give your answer correct to the nearest foot. (c) How far from the firing point will the projectile strike the ground? Give your answer correct to the nearest foot.

Solution

h(x) = (-32x²_/96²) + x

comparing with y =ax²+bx +c

a) distance  from the firing point to the point where the height of the projectile a maximum is the x coordinate of vertex of parabola

x =-b/(2a)

x=-1/(2 (-32_/96²) )

x=96²/64

x=144

144 feet  from the firing point the height of the projectile a maximum

b)maximum height h(144)=(-32*144²_/96²) + 144

maximum height =72 feet

c)when projectile strike the ground h(x)=0

(-32x²_/96²) + x =0

(-32x_/96²) + 1=0

(32x_/96²) =1

x=96²_/32

x=288

288feet from the firing point the projectile strikes the ground