Question

Find the horizontal distance of a soccer ball being kicked at a 40 degree angle with a velocity of 50 feet per second, to the nearest foot

Answer 1

Answer:

77 ft

Step-by-step explanation:

The horizontal distance is the range and it is a projectile motion.

So, range R = u²sin2θ/g where u = velocity = 50 ft/s, g = acceleration due to gravity = 32 ft/s² and θ = projection angle = 40°.

Substituting the values of the variables into the equation, we have

R = u²sin2θ/g

= (50 ft/s)²sin2(40°)/32 ft/s²

= (2500 ft²/s²)sin80°/32 ft/s²

= (2500 ft²/s²) × 0.9848/32 ft/s²

= 2462.02 ft²/s² ÷ 32 ft/s²

= 76.94 ft

≅ 77 ft to the nearest foot