A long 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.

He jumps 4.37 m

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