A projectile is launched at an angle of 54.6 degrees above the horizontal with an initial speed of 225 m/s and lands at the same level from which it was launched. The range is 205 meters. The time of flight is how many seconds?
To find the time of flight of the projectile, we can use the horizontal component of the initial velocity and the range of the projectile.
The horizontal component of the initial velocity (Vx) can be calculated using the initial speed (Vi) and the launch angle (θ). It is given by the equation:
Vx = Vi * cos(θ)
In this case, Vi = 225 m/s and θ = 54.6 degrees. Let's calculate Vx:
Vx = 225 * cos(54.6)
Vx ≈ 225 * 0.5878
Vx ≈ 132.28 m/s
Now, we can use the horizontal component of the velocity and the range to find the time of flight (T). The formula to calculate the time of flight is:
T = Range / Vx
Substituting the given values into the equation:
T = 205 / 132.28
T ≈ 1.55 seconds
Therefore, the time of flight of the projectile is approximately 1.55 seconds.