Answer: 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 [tex]a(t) = (0, 32.2\frac{ft}{s^{2} } )[/tex] 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: [tex]v(t) = (450*cos(30)\frac{ft}{s} , 450*sin(30)\frac{ft}{s} + 32.2\frac{ft}{s^{2} }*t )[/tex]
and using that the initial position is (0,0) we integrate the velocity over time to obtain the position:[tex]r(t) (450*cos(30)\frac{ft}{s}*t , 450*sin(30)\frac{ft}{s}*t + 32.2\frac{ft}{s^{2} }*\frac{t^{2} }{2} )[/tex]
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.
