A Place Kicker Kicks A Football With The Velocity Of 20 M/S And At Angle Of 553 \Xb0 How High It Travels ?

A place kicker kicks a football with the velocity of 20 m/s and at angle of 553 ° how high it travels ?

 

Given:

angle theta= 20 m/s

Vi= 553 degrees

Required: height at the highest/maximum point

Solution:

Viy= (sin(angle theta))(Vi)

    = (sin 553 degrees)(20m/s)

    = -4.50

time at the highest point= Vy-Viy/g

                                            = 0-(-4.50)/9.8m/s^2

                                            = 4.5/9.8 m/s^2

                                            = 0.46 s

total time= 2(time at the highest point)

               = 2(0.46 s)

               = 0.92 s

height= Viy(time at highest point)+g(time at highest point)^2/2

          = (-2.07)+(2.25)

          = -4.66 m


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