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An object is launched with an initial velocity of 36.3 m/s at an angle of 57.7 degrees relative to the +x direction. If it is initially at ground level, what amount of time does it take to return to the ground?

Answer :

Answer:

Explanation:

To find the amount of time it takes for the object to return to the ground, we can analyze the vertical motion of the object.

Given:

Initial velocity (v₀) = 36.3 m/s

Launch angle (θ) = 57.7 degrees

We can break down the initial velocity into its horizontal and vertical components:

v₀x = v₀ * cos(θ)

v₀y = v₀ * sin(θ)

Since the object is launched at ground level, the initial vertical position (y₀) is 0.

The equation for vertical displacement (y) can be expressed as:

y = y₀ + v₀y * t - (1/2) * g * t²

where:

y is the vertical displacement at time t,

v₀y is the vertical component of the initial velocity,

g is the acceleration due to gravity (-9.8 m/s²), and

t is the time.

The object will return to the ground when its vertical displacement is 0. So we can set y = 0 and solve for t.

0 = v₀y * t - (1/2) * g * t²

Rearranging the equation:

(1/2) * g * t² = v₀y * t

Simplifying:

(1/2) * g * t = v₀y

t = (2 * v₀y) / g

Substituting the values:

t = (2 * v₀ * sin(θ)) / g

t = (2 * 36.3 m/s * sin(57.7°)) / 9.8 m/s²

Calculating this expression will give us the amount of time it takes for the object to return to the ground.

know more acceleration: brainly.com/question/2303856

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