An archer fires an arrow at an angle of 9° above the horizontal with a resultant velocity of 24 m/s

Explanation:

Here is the complete question. An archer fires an arrow at an angle of 9° above the horizontal with a resultant velocity of 24 m/s, How high will it go? What horizontal distance will it travel?

To determine the maximum height it will go, we will use the formula for calculating the maximum height as shown;

H = u²sin²theta/2g

u is the velocity = 24m/s

theta is angle of launch = 9°

g is acceleration due to gravity = 9.8m/s²

Substitute

H = 24²(sin9°)²/2(9.8)

H = 576×0.0245/19.6

H = 14.0957/19.6

H = 0.719m

Hence the arrow can travel 0.719m High

The horizontal distance it will travel is known as RANGE. This is expressed as;

R = U²sin2(theta)/g

R = 24²sin18°/9.8

R = 576sin18°/9.8

R = 177.99/9.8

R = 18.16m

Hence the horizontal distance the arrow can travel is 18.16m

the answer is a. at the equator

its c i just checked it