Along jumper leaves the ground at an angle of 22.5 ∘ measured from the horizontal, and at a speed of 11.0 m/s. how far does he jump? hint: first determine how long he is in the air, which is twice the time it takes him to reach his maximum height.

Explanation:

Consider the vertical motion of long jumper

We have equation of motion v = u + at

     Initial velocity, u = 11 sin 22.5 = 4.21 m/s

     Final velocity, v =  0 m/s    

     Acceleration, a = -9.81 m/s²

     Substituting

                      v = u + at  

                      0 = 4.21 + -9.81 x t

                      t = 0.43 s

So the Long jumper is in air for 0.43 seconds.

Now consider the vertical motion of long jumper

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 11 cos 22.5 = 10.16 m/s

        Acceleration, a = 0 m/s²  

        Time, t = 0.43 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 10.16 x 0.43 + 0.5 x 0 x 0.43²

                      s = 4.37 m

He jumps 4.37 m

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.

the answer is b ) square root of 5