Question

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

Answer 1

A projectile is fired with an initial velocity of 450 ft / sec and with an elevation angle of 30 ° above the horizontal, calculate a) its position and speed after 8s,
The position after 8 seconds will be as follows
horizontal distance:
x=V×t
x=450×8
=3600 ft

 b) the maximum height reached will be given by:
H=[v² sin²θ]/2g
Plugging in the values we get:
H=[450²×sin²30]/(2×10)
H=2,531.25 ft

Answer 2

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 $a(t) = (0, 32.2\frac{ft}{s^{2} } )$ 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: $v(t) = (450*cos(30)\frac{ft}{s} , 450*sin(30)\frac{ft}{s} + 32.2\frac{ft}{s^{2} }*t )$

and using that the initial position is (0,0) we integrate the velocity over time to obtain the position:$r(t) (450*cos(30)\frac{ft}{s}*t , 450*sin(30)\frac{ft}{s}*t + 32.2\frac{ft}{s^{2} }*\frac{t^{2} }{2} )$

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.