Answer:
3.53 s
Explanation:
The projectile is launched with a horizontal speed Vx = 30 m/s. This is a component of the speed vector of the projectile. The angle is 30 degrees, so we can calculate the vertical component:
tg(a) = Vy/Vx
Vy = tg(a) * Vx
Vy = tg(30) * 30 = 17.3 m/s
Now, since the projectile is at free falit is only affected by the acceleration of gravity, therefore we can say it is at constant acceleration and we can use this equation:
Y(t) = Y0 + Vy0 * t * 1/2 * a * t^2
In this case the projectile is shot from the ground, so Y0 = 0.
a is the gravity, -9.81 m/s^2 (negative because it points down)
So we end up with
Y(0) = 0 + 17.3 * t + 1/2 * (-9.81) * t^2
If we equal this to zero we can find the moments it is at zero height, these will be the moment it was shot (t=0) and the moment it hit the ground. The difference between these is the time it spent on the air.
0 = 17.3 * t - 4.9 * t^2
0 = t * (17.3 - 4.9 * t)
t = 0 is one of the solutions as expected
0 = 17.3 - 4.9 * t
4.9 * t = 17.3
t = 17.3/4.9 = 3.53 s
This is the time when it hit the ground, then
3.53 - 0 = 3.53 s
This is the time it spend on the air
