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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?

Respuesta :

Answer:

H = 4.12 m

Explanation:

As we know that horizontal range is the distance moved in horizontal direction

Since horizontal direction has no acceleration

so here we have

[tex]Range = v_x T[/tex]

here we know that

[tex]v_x = vcos32[/tex]

so from above formula

[tex]92 = (vcos32)(6.4)[/tex]

[tex]v = 16.95 m/s[/tex]

now we have maximum height is given as

[tex]H = \frac{(vsin32)^2}{2g}[/tex]

[tex]H = \frac{(16.95 sin32)^2}{2(9.8)}[/tex]

[tex]H = 4.12 m[/tex]

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