For this problem we will use the constant acceleration equations.
1. Flight time:
We will use the equation
y = y0 + V0y t + (1/2) a t^2 --> Italicized means sub
where y0 is the initial y position and V0y is the initial y component of velocity.
Flight time means that we take off from the ground (y0 = 0) and land on the ground (y = 0). We can substitute that into our equation:
0 = 0 + V0y t + (1/2) a t^2
To find the initial y component of velocity, we multiple the initial velocity by sine of theta (angle off the horizontal). So V0y = 12*sin(30) = 6.
To find the acceleration, we assume that all this is taking place on Earth so we use the acceleration of gravity which is (-9.8m/s^2).
Using all this info we plug into the formula:
0 = 6t - 4.5t^2
t (6-4.8t) = 0
t= 0 s
or
t = 6/4.8 = 1.25 s. The reason we get two answers is because the equation gives us the time stamps of when the projectile was on the ground, which is before it was thrown and when it landed. Therefore, the flight time is 1.25 s.
Max height:
To find the max height we first ned to find the total distance traveled, which is x.
Using 1.25 s, we simple plug into our previous formula:
x-x0 = V0xt + (1/2)axt^2
x-0 = (12*cos(30)) * (1.25) + (1/2) * (0) * (1.25^2)
x = 13m
The reason ax = 0 is because there is no acceleration in the x direction. Only the y direction is affected because gravity points downwards AND we are assuming there is no air resistance.
Now that we have 13m:
plug in half of total distance (position where projectile is at max height) and solve for t again:
(13/2) = (12*cos(30)) * t + 0
t = (13/2)/(12cos(30)) = 0.625 s
Hope that helps man
