A projectile is fired horizontally from a gun that is 45.0m above flat ground, emerging from the gun with a speed of250m/s. Calculate the amount of time taken by the projectile in the air?
time:
h=1/2 g t^2
t= sqrt(2*45/9.8) seconds
horizontal velocity has nothing to do with it.
To calculate the amount of time taken by the projectile in the air, we can use the kinematic equation for vertical motion:
Δy = Vyi * t + (1/2) * a * t^2
Where:
Δy is the vertical displacement (45.0 m)
Vyi is the initial vertical velocity (0 m/s, as the projectile is fired horizontally)
a is the acceleration due to gravity (-9.8 m/s^2)
t is the time taken
Since the initial vertical velocity is 0 m/s, the first term of the equation cancels out, leaving us with:
Δy = (1/2) * a * t^2
Rearranging the equation to isolate t:
t^2 = (2 * Δy) / a
Substituting the values:
t^2 = (2 * 45.0 m) / (-9.8 m/s^2)
t^2 = -9.18
(Note: Since time cannot be negative, we can ignore the negative sign)
Taking the square root of both sides:
t ≈ 3.03 seconds
Therefore, the projectile will be in the air for approximately 3.03 seconds.