I have some tricky problems i could not get by

Question

A projectile is fired at an inclination of 45° to the horizontal, with a muzzle velocity of 400 feet per

second. The height h of the projectile is modeled by

h(x) = -32x^2/(400)^2+x

where x is the horizontal distance of the projectile from the firing point.

a. Find the maximum height of the projectile.

b. When the height of the projectile is 800 feet above the ground, how far has the projectile

traveled horizontally from the firing point?

Answer 1

That's not the formula for the height of a projectile. The formula is h(x) = -32x² + sin(45^o)400x. The sin(45^o) = √2 / 2 = .7071. So, the equation is h(x) = -32x² + 283x. BTW, the variable x is time. So, using t makes it easier to understand, h(t) = -32t² + 283t.

To figure out the maximum height, you first figure out the time it takes for the projectile to both go up and come back down to ground. The height at the ground is zero, so h(t) = 0 = -32t² + 283t = t * (-32t + 283) One answer is zero. (The projectile starts on the ground!) The other answer is 8.8 seconds to hit the ground again.

The project takes the same amount of time to up as it does to go down, so the maximum height is at half that time, 4.4 seconds. So h(4.4) = -32(4.4)^2+283(4.4) = 626 feet high.

The problem with the second half of the question is that the projectile never gets to 800 feet. But if you want to figure how far it's gone at some other height, you figure out the time it gets that height and plug that time into the formula for the horizontal distance d(t) = cos(45^o)400t. That's d(t) = 283t. So, for example, the projectile would land 283(8.8) = 2490 feet away.