A projectile is fired with an initial speed of 230 m/s and an angle of elevation 60°. The projectile is fired from a position 100 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.
(b) Find the maximum height reached.
(c) Find the speed at impact.

Respuesta :

The range of the projectile is 4675 m and the maximum height is 2024 m.

A projectile is the motion of an object which is thrown into space and subjected to the acceleration due to gravity.

Given that:

Initial speed (u) = 230 m/s, acceleration due to gravity (g) = 9.8 m/s², angle (θ) = 60°

a)

[tex]Range=\frac{u^2sin2\theta}{g} =\frac{230^2*sin(2*60)}{9.8} =4675\ m[/tex]

b)

[tex]Maximum\ height(h)=\frac{u^2sin^2\theta}{2g} =\frac{230^2*sin^2(60)}{2*9.8} =2024\ m[/tex]

c)

[tex]Speed\ at\ impact=\sqrt{2gh}=\sqrt{2*9.8*2024} =199\ m/s[/tex]

The range of the projectile is 4675 m and the maximum height is 2024 m.

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