Un proyectil se dispara con una velocidad inicial de 450 pies/ seg y con un angulo de elevacion de 30° sobre la horizontal, calcular a) su posición y velocidad despues de 8s, b) la altura máxima

This says: A projectile is fired with initial velocity of 450 feet per second and at 30° respect with the ground.

If the initial position is (0,0) then we can start to derive the movement equations. Where the notation used is (x,y)

The acceleration is because the only force acting on the projectile is the gravity in the y axis.

for the velocity we integrate can see that the initial velocity will be 450*cos(30°) feet per second in the x axis, and 450*sin(30°) feet per second in the y axis.

then the velocity, integrating over time the acceleration and adding the initial velocity is:

and using that the initial position is (0,0) we integrate the velocity over time to obtain the position:

a) here we want to know v(8s) and r(8s)

if you put t= 8s in the equations you get v(8s) = (389.71, -32.6) in feet per second.  r(8s) = (3117.68, 769.6) in feets.

b) the maximum height. Here we need to se when the velocity in Y is 0, and put that time in the position equation.

so the velocity in y is 450*sin(30) - 32.2*t, if we igualate it to zero, we get that t = 350*sin(30)/32.2 = 6.99s

so the maximum height is reached at the time 6.99 seconds. putting it in the position equation for y we get ymax = 786.1 feet.

not sure but it may be b, as i said i'm not sure.

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