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A projectile is fired with an initial speed of 170 m/s and angle of elevation 60°. The projectile is fired from a position 120 m above the ground. (Recall g ≈ 9.8 m/s2. Round your answers to the nearest whole number.) (a) Find the range of the projectile. 4590 Incorrect: Your answer is incorrect. m (b) Find the maximum height reached. 1106 Incorrect: Your answer is incorrect. m (c) Find the speed at impact.

Respuesta :

Answer:

a)R=2551.28 m

b)h=1104.74 m

c)V=48.52 m/s

Explanation:

Given

u=170 m/s

θ=60°

Height = 120 m

Range ,R

[tex]R=\dfrac{u^2sin2\theta }{g}[/tex]

[tex]R=\dfrac{170^2\times sin120 }{9.81}[/tex]

R=2551.28 m

Maximum height,h

[tex]h=\dfrac{u^2sin^2\theta }{2g}[/tex]

[tex]h=\dfrac{170^2sin^260 }{2\times 9.81}[/tex]

h=1104.74 m

The speed at impact

[tex]V=\sqrt{2gH}[/tex]

[tex]V=\sqrt{2\times 9.81\times 120}\ m/s[/tex]

V=48.52 m/s

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