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A rocket is launched at an angle of [tex]51.2^\circ[/tex] with a velocity of [tex]453 \, \text{m/s}[/tex] and hits the ground.

(a) What is the height of the rocket at [tex]35.9 \, \text{s}[/tex]?

Answer :

Final answer:

The height of the rocket at 35.9 s can be calculated using the equation for vertical motion.

Explanation:

The height of the rocket at 35.9 s can be calculated using the equation for vertical motion. Since we know the initial vertical velocity (v₀ = 453 m/s), the time of flight (t = 35.9 s), and the acceleration due to gravity (g = 9.8 m/s²), we can use the formula:

Δy = (v₀ * t) + (1/2 * g * t²)

Plugging in the values, we get:

Δy = (453 * 35.9) + (1/2 * 9.8 * 35.9²)

Simplifying, we find that the height of the rocket at 35.9 s is approximately 256,788.15 m.

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