Question

Linda shoots an arrow at a target in an archery competition. The arc of the arrow can be modeled by the equation y= -0.02x to the power of 2 + 0.65+4 where x is the horizontal distance (in meters) from Linda and y is the height (in meters) of the arrow. How far from Linda does the arrow hit the ground? Round to the nearest tenth.

Answer 1

37.8 m

Explanation:

Given:-

- The arc trajectory of the arrow is modeled by:

y = -0.02x^2 + 0.65x + 4

Where, x is the horizontal distance (in meters) from Linda

y is the height (in meters) of the arrow

Find:-

How far from Linda does the arrow hit the ground? Round to the nearest tenth.

Solution:-

- We are to determine the range of the projectile trajectory of the arrow. The maximum distance "x_max" occurs when the arrow hits the ground.

- Set the trajectory height of arrow from linda , y = 0:

0 = -0.02x^2 + 0.65x + 4

- Solve the quadratic equation:

x = -5.29 m , x = 37.8 m

- The negative distance x lies at the back of Linda and hence can be ignored. The maximum distance travelled by the arrow would be = 37.8 m

Answer 2

37.8metres

Explanation:

The arc of the arrow can be modeled by the equation:

y=-0.02x²+0.65x+4

Where x is the horizontal distance (in meters) from Linda and y is the height (in meters) of the arrow.

The arrow hits the ground when its height (y) is zero.

Therefore, we determine the value(s) of x for which:

y=-0.02x²+0.65x+4=0

Using a calculator to solve the quadratic equation:

x=37.79 or -5.29

Since the distance cannot be a negative value, we ignore -5.29.

The distance from Linda when the arrow hits the ground is 37.8metres (to the nearest tenth)