Aprojectile is launched at an angle of 30 degrees with the horizontal and a speed of 30 m/s. how much time does it spend in the air?
a.)
2.7 s
b.)
1.5 s
c.)
3.1 s
d.)
1.8 s

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

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because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14