y=(v/u)x+(g)/(2(u^2))*x^2

where

y is the object's vertical distance (in feet) above from where it was thrown

x is the object's horizontal distance (in feet) away from where it was thrown

and

u is the objects' initial horizontal velocity

v is the object's initial vertical velocity

g is the force exerted by gravity.

Suppose a ball is launched from ground level (on level ground) with an initial horizontal velocity of 50 feet/second and an initial vertical velocity of 100 feet/second and that the force due to gravity was -16 feet/second2

a) Substitute those values into the given formula and simplify.

b) Find how far the ball is away when the ball hits the ground again. Show the equation you solved and tell or show how you solved it.

a) To substitute the given values into the formula and simplify:

y = (v/u)x + (g/(2u^2))x^2

Given:
u = 50 feet/second
v = 100 feet/second
g = -16 feet/second^2

Substituting the given values into the formula:

y = (100/50)x + (-16/(2(50^2)))x^2

Simplify:

y = 2x - 0.016x^2

b) To find how far the ball is away when it hits the ground again (when y = 0), we need to set y = 0 and solve the resulting quadratic equation.

Setting y = 0:

0 = 2x - 0.016x^2

Rearranging the equation to standard quadratic form:

0.016x^2 - 2x = 0

Simplifying, we can divide the equation by 0.016:

x^2 - 125x = 0

Factoring out an x:

x(x - 125) = 0

Setting each factor equal to zero and solving for x:

1) x = 0
2) x - 125 = 0

Solving for x:

1) x = 0
2) x = 125

Therefore, the ball is away 125 feet when it hits the ground again.

a) To substitute the given values into the formula and simplify, we need to replace the variables with their corresponding values.

Given:
u = 50 feet/second (initial horizontal velocity)
v = 100 feet/second (initial vertical velocity)
g = -16 feet/second^2 (force due to gravity)

Substituting the values into the formula:

y = (v/u)x + (g/(2u^2))x^2

y = (100/50)x + (-16/(2 * 50^2))x^2

Simplifying further:

y = 2x - (16/5000)x^2

Therefore, the simplified equation is y = 2x - (16/5000)x^2.

b) To find how far the ball is away when it hits the ground again, we need to determine the value of x when y = 0. This is because when the ball hits the ground, its vertical distance (y) is zero.

Setting y = 0 in the equation:

0 = 2x - (16/5000)x^2

Now, we need to solve this quadratic equation for x. There are multiple ways to solve it, but let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For ease of calculation, let's simplify the equation:

0 = (16/5000)x^2 - 2x

Comparing with the general quadratic equation form ax^2 + bx + c = 0, we have:

a = 16/5000
b = -2
c = 0

Plugging these values into the quadratic formula:

x = (-(-2) ± √((-2)^2 - 4 * (16/5000) * 0)) / (2 * (16/5000))

Simplifying further:

x = (2 ± √(4 - 0)) / (32/5000)

x = (2 ± √4) / (32/5000)

x = (2 ± 2) / (32/5000)

x = 4/ (32/5000) or 0/ (32/5000)

x = 625 feet or 0 feet

Since the ball is launched horizontally with a velocity of 50 feet/second, and the force due to gravity is negative (indicating downward acceleration), the ball will hit the ground again at x = 625 feet.