how long will an object remain in the air if it is launched with an initial velocity of 30.0 m/s [65.0 degrees]
To determine the total time an object remains in the air when launched with an initial velocity, we can break down the initial velocity into its horizontal and vertical components.
The vertical component of the initial velocity can be calculated as:
Vy = V * sin(theta)
Vy = 30.0 m/s * sin(65.0 degrees)
Vy ≈ 30.0 m/s * 0.9063
Vy ≈ 27.187 m/s
The acceleration due to gravity is -9.81 m/s^2 (negative because it is acting downwards).
To calculate the total time the object remains in the air, we can use the following kinematic equation:
s = Vyi * t + 0.5 * a * t^2
Where:
s = displacement in the vertical direction (equals 0 at the highest point)
Vyi = initial vertical velocity (27.187 m/s)
a = acceleration due to gravity (-9.81 m/s^2)
t = time in seconds
Rearranging the equation to solve for time:
0 = 27.187 * t + 0.5 * (-9.81) * t^2
0 = 27.187t - 4.905t^2
This is a quadratic equation, and we need to solve it for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Plugging in the values:
t = (-27.187 ± sqrt(27.187^2 - 4 * -4.905 * 0)) / 2 * -4.905
t = (-27.187 ± sqrt(739.868569 - 0)) / -9.81
t = (-27.187 ± √739.868569) / -9.81
t = (-27.187 ± 27.197) / -9.81
Taking the positive value for t:
t ≈ -0.01 s (this negative value can be ignored)
Therefore, the object will remain in the air for approximately 5.514 seconds.