Posted by **Geoff** on Friday, July 19, 2013 at 10:17pm.

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.

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