Abullet is shot at an angle of 32° above the horizontal on a level surface. it travels in the air for 6.4 seconds before it strikes the ground 92m from the shooter. what was the maximum height reached by the bullet? round to one decimal place and include units.

Maximum height reached by the bullet = 4.01 m

Explanation:

Horizontal displacement = 92 m

Time taken = 6.4 s

Horizontal velocity

We have angle of projection = 32°

Horizontal velocity = u cos 32 = 14.375

                      u = 16.95 m/s

Vertical velocity = u sin θ = 16.95 x sin 32 = 8.98 m/s

Time of flight till it reaches maximum height = 0.5 x 6.4 = 3.2s

Now we have vertical motion of bullet

        S = ut + 0.5 at²

       Vertical velocity = u = 16.95 m/s

       a = -9.81 m/s²

        t = 3.2s

Substituting

        S = 16.95 x 3.2 - 0.5 x 9.81 x 3.2² = 4.01 m

Maximum height reached by the bullet = 4.01 m

not really sure but thinking 1.54 mm/us

the higher the current the higher the mag field around it the higher the influence on the compass