In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0580 s, during which time it experiences an acceleration of 376 m/s2. The ball is launched at an angle of 59.9° above the ground. Determine the (a) horizontal and (b) vertical components of the launch velocity.

V₀ₓ = 10.94 m/s

V₀y = 18.87 m/s

Explanation:

To find the launch velocity, we use 1st equation of motion.

Vf = Vi + at

where,

Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

a = acceleration = 376 m/s²

t = time = 0.058 s

Therefore,

V₀ = 0 m/s + (376 m/s²)(0.058 s)

V₀ = 21.81 m/s

Now, for x-component:

V₀ₓ = V₀ Cos θ

where,

V₀ₓ = x-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀ₓ = (21.81 m/s)(Cos 59.9°)

V₀ₓ = 10.94 m/s

for y-component:

V₀ₓ = V₀ Sin θ

where,

V₀y = y-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀y = (21.81 m/s)(Sin 59.9°)

V₀y = 18.87 m/s

we need to know the time interval during which the force acts. impulse of force = momentum change