Question

A football is kicked from a tee at an angle of 53 degrees above the horizontal with an initial speed of 20 m/s. A coach times the flight of the ball and finds it was in the air for 3.3 seconds. When the football is at its maximum vertical height its velocity and acceleration are (neglect air resistance)?

Answer 1

v = 12.036 m/s, horizontal

g = 9.8 m/s², vertical downward

Step-by-step explanation:

The ball describes a parabolic path, and the equations of the movement are:

Equation of the uniform rectilinear motion (horizontal ) :

x = vx*t : Equation (1)

Equations of the uniformly accelerated rectilinear motion of upward (vertical ) :

y = (v₀y)*t - (1/2)*g*t² Equation (2)

vfy² = v₀y² -2gy Equation (3)

vfy = v₀y -gt Equation (4)

Where:

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity in m

y: vertical position in meters (m)

v₀y: initial vertical velocity in m/s

vfy: final vertical velocity in m/s

g: acceleration due to gravity in m/s²

Known data

v₀ = 20 m/s , at an angle α=53° ,above the horizontal

Flight time = 3.3 s

g = 9.8 m/s²

Speed of the ball (v) in the maximum height

vy=0 : vertical speed

vx = (v₀) (cos α) = (20 m/s)*(cos 53°) : horizontal speed

vx = 12.036 m/s

v = vx = 12.036 m/s , horizontal

Acceleration (g) of the ball (v) in the maximum height

g = -9.8 m/s²

g = 9.8 m/s², vertical downward