How far from Jacob does the ball land ?

Question

The arc of the ball can be modeled by the equation y=-0.05x squared+0.75x+4,x is the horizontal distance (in meters),y is the height( in meters) of the ball.

Answer 1

You are given the equation: 

 

y = -0.05x² + 0.75x + 4

 

 

If a ball is thrown from ground level, when it lands, it will land back on the ground. We can say that our height (y) will equal 0 

 

0 = -0.05x² + 0.75x + 4

 

 

 

the equation is in the form of y = ax² + bx + c, so you can  use the quadratic equation to solve for the values of x

 

the quadratic equation is:  

 

 

x = (-b ± sqrt (b² -4ac)) / (2a)

 

plugging in the values, we get:

 

x = (-0.75 ± sqrt (0.75² -4*(-.05)*(4)) / (2*(-.05))

 

simplifying the equation, we get

 

x = (-0.75 ± sqrt (0.5625 + 0.8)) / (-0.1)

x = (-0.75 ± sqrt (1.3625) / (-0.1)

x = (-0.75 ± 1.167) / (-0.1)

 

we have a ± so there will be two answers for x

 

x = (-0.75 + 1.67) / (-0.1)

 

x = 19.17

 

AND 

 

x = (-0.75 - 1.67) / (-0.1)

 

x = -4.17

 

the values of x gives use the distance the ball traveled 

 

since it doesn't make sense for the ball to travel a negative distance we can say the ball lands 19.17 meters from Jacob