Question

A projectile is fired from a cliff 220 ft above water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 65 ft per secound. the height ,h, of the projectile abore water is given, h(x)=(-32x^2)/ ((65)^2 ) +x+220. x is the horizontal distance of the projectile from the face of the cliff. What is the maximum value of x?

Answer 1

248.79 ft

Explanation:

A projectile is fired from a cliff 220 ft above water at an inclination of 45 degrees to the horizontal, with a muzzle velocity of 65 ft per second.

`h(x)=-(32x^2)/(65^2)+x+220`

For maximum value of x, h(x)≥0

`-(32x^2)/(65^2)+x+220\geq0`

Solve quadratic equation for x

`-(1)/(4225)(32x^2-4225x-929500)\geq0`

`32x^2-4225x-929500\leq0`

Using quadratic formula,

`x=(-b\pm√(b^2-4ac))/(2a)`

where, a=32, b=-4225, c=-929500

`x=(4225\pm√(4225^2-4(32)(-929500)))/(2(32))`

`x\geq-116.75\text{ and }x\leq 248.79`

Hence, The maximum value of x will be 248.79 ft