Answer:
To find the time it takes for the projectile to return to the ground at the same height, we can analyze the vertical motion of the projectile. The horizontal motion does not affect the time of flight in this case.
We can break down the initial velocity of the projectile into its horizontal and vertical components. The horizontal component remains constant throughout the motion, while the vertical component is affected by gravity.
Given:
Initial speed (v₀) = 32 m/s
Launch angle (θ) = 46°
First, we can find the vertical component of the initial velocity (v₀ₓ) using trigonometry:
v₀ₓ = v₀ * cos(θ)
v₀ₓ = 32 * cos(46°)
v₀ₓ ≈ 32 * 0.7193
v₀ₓ ≈ 23.02 m/s (rounded to two decimal places)
The time taken for the projectile to reach its highest point (t₁) can be calculated using the formula:
t₁ = v₀ₓ / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
t₁ = 23.02 / 9.8
t₁ ≈ 2.35 seconds (rounded to two decimal places)
Since the time taken to reach the highest point is the same as the time taken to descend from the highest point to the ground, the total time of flight is:
t_total = 2 * t₁
t_total ≈ 2 * 2.35
t_total ≈ 4.70 seconds (rounded to two decimal places)
Therefore, the time between the projectile leaving the ground and returning to the ground at the same height is approximately 4.70 seconds.