A projectile is shot from the ground at a velocity of #29 m/s# and at an angle of #(2pi)/3#. How long will it take for the projectile to land?

1 Answer
Jun 10, 2016

Answer:

#x=74.24" m"#

Explanation:

enter image source here

#"The velocity of projectile can be split two components as vertical and horizontal"#

#"the vertical component can be calculated using the formula:"#
#v_y=v_i*sin alpha-g*t" "figure" "1#

enter image source here

#"you are seeing how the vertical component of velocity is changing "#

#"please note that the vertical component of velocity " v_y" is zero at maximum height"#

#0=v_i*sin alpha-g*t#
#g*t=v_i*sin alpha#
#t=(v_i*sin alpha)/g" time elapsed to the maximum height"#

#"2*t gives us the traveling time"#

#color(red)(t_t=(2*v_i*sin alpha)/g)#

enter image source here

#"the object is conveyed by the horizontal component of velocity "v_x" figure 3"#

#"the horizontal component of velocity is calculated by:"#

#color(green)(v_x=v_i*cos alpha)#
#"the horizontal component of velocity doesn't change"#

#"and doesn't depend on time"#

enter image source here

#"the horizontal position of object can be calculated by:"#

#x=color(green)(v_x)*color(red)(t)#

#x=color(green)(v_i*cos alpha)*color(red)(t)#

#"now; we can write "color(red)(t=t_t)#

#x=(v_i*cos alpha*(2*v_i*sin alpha))/g#

#"so "2*sin alpha*cos alpha=sin(2 alpha)#

#x=(v_i^2*sin(2alpha))/g#

#"where "v_i=29" m/s"" "alpha=((2pi)/3) #

#x=(29^2*sin(2*(2pi)/3))/(9.81)#

#x=(841*0.866)/(9.81)#

#x=74.24" m"#