A projectile is shot at an angle of pi/6 and a velocity of 3 2 m/s. How far away will the projectile land?

1 Answer
Jul 12, 2016

x_m=90.40" "m

Explanation:

enter image source here
"please look over the animation carefully"
"The initial velocity of object can be split into two components"
"as vertical and horizontal. "(v_x,v_y)

v_x=v_i*cos alpha
v_y=v_i*sin alpha-g*t

enter image source here

"we can find the time elapsed using the component "v_y

"we can write "v_y=0 " for the maximum height."

v_y=v_i*sin alpha-g*t=0

v_i*sin alpha=g*t

t=(v_i*sin alpha)/g

t_f:"represents the flaying time"

t_f=2*t=(2*v_i*sin alpha)/g

enter image source here

x_m=v_x*v_f

x_m=(v_i*cos alpha*2*v_i*sin alpha)/g

x_m=(v_i^2*2*sin alpha*cos alpha)/g

2*sin alpha*cos alpha=sin 2 alpha

x_m=(v_i^2*sin 2alpha)/g

"where:"
v_i=32" "m/s
2*alpha=pi/6*2=pi/3
g=9.81" "m/s^2

x_m=(32^2*0.866)/(9.81)

x_m=90.40" "m