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A bullet 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.

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Blacklash

Answer:

Maximum height reached by the bullet = 4.01 m

Explanation:

Horizontal displacement = 92 m

Time taken = 6.4 s

Horizontal velocity

           [tex]=\frac{92}{6.4}=14.375m/s[/tex]

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

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