Question

A football is kicked 67.3 meters. If the ball is in the air 5.37s, with what initial velocity was it kicked?

magnitude?
direction? above the horizontal

Answer 1

Answer 2

We can use the kinematic equation of motion for a projectile to solve this problem:

Δy = vi * t + (1/2) * a * t^2

where Δy is the vertical displacement (which we can assume is zero since the ball returns to its initial height), vi is the initial velocity, t is the time of flight, and a is the acceleration due to gravity (-9.81 m/s^2).

Since the ball is kicked horizontally, there is no initial vertical velocity. Thus, the only initial velocity component is the horizontal velocity, which remains constant throughout the motion.

The horizontal displacement can be found using the formula:

Δx = vix * t

where Δx is the horizontal displacement, and vix is the initial horizontal velocity.

Substituting the given values:

Δy = 0 t = 5.37 s a = -9.81 m/s^2 Δx = 67.3 m

Solving for vix:

Δx = vix * t vix = Δx / t

vix = 67.3 m / 5.37 s vix = 12.525 m/s

Therefore, the initial horizontal velocity was 12.525 m/s.

To find the magnitude of the initial velocity, we can use the Pythagorean theorem, which states that the magnitude of the initial velocity is equal to the square root of the sum of the squares of the horizontal and vertical components of the initial velocity.

Since there is no initial vertical velocity, the magnitude of the initial velocity is equal to the magnitude of the initial horizontal velocity, which is 12.525 m/s.

To find the direction of the initial velocity above the horizontal, we can use the trigonometric formula:

θ = arctan(viy / vix)

where θ is the angle above the horizontal, viy is the initial vertical velocity (which is zero), and vix is the initial horizontal velocity.

θ = arctan(0 / 12.525) θ = 0°

Therefore, the direction of the initial velocity above the horizontal is 0° (i.e., it was kicked horizontally).