Question

A soccer ball is kicked from the playing field at a 45° angle. If the ball is in the air for 3 seconds, what is the maximum height achieved by the ball?

Answer 1

The maximum height achieved by the soccer ball kicked at a 45° angle can be calculated using the equations of projectile motion. The initial velocity can be found using the time of flight and launch angle, and then substituted into the equation for maximum height. The maximum height is approximately 21.1 m.

Step-by-step explanation:

The maximum height achieved by the soccer ball can be determined using the equations of projectile motion. The vertical motion of the ball can be described by the equation:

h = (v₀sinθ)² / (2g)

where h is the maximum height, v₀ is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. In this case, the angle of launch is 45° and the time of flight is given to be 3 seconds. Using this information, we can find the initial velocity:

v₀ = gt / sinθ = (9.8 m/s²)(3 s) / sin(45°) ≈ 20.7 m/s

Substituting this value into the equation for maximum height:

h = (20.7 m/s * sin(45°))^2 / (2(9.8 m/s²)) ≈ 21.1 m