A bullet has a speed of 250 m/s as it leaves a rifle. If the rifle is fired 50 degrees from the ground a. What is the time flight in the ground? b. What is the maximum height? c. What is the range?

1 Answer
Feb 16, 2018

#a. 39.08 " seconds"#
#b. 1871 " meter"#
#c. 6280 " meter"#

Explanation:

#v_x = 250 * cos(50°) = 160.697 m/s#
#v_y = 250 * sin(50°) = 191.511 m/s#
#v_y = g * t_{fall}#
#=> t_{fall} = v_y / g = 191.511 / 9.8 = 19.54 s#
#=> t_{flight} = 2*t_{fall} = 39.08 s#
#h = g*t_{fall}^2/2 = 1871 m#
#"range "= v_x * t_{flight} = 160.697 * 39.08 = 6280 m#

#"with"#
#g = " gravity constant = 9.8 m/s²"#
#v_x = " horizontal component of the initial velocity"#
#v_y = " vertical component of the initial velocity"#
#h = " height in meter (m)"#
#t_{fall} = " time to fall from highest point to the ground in sec."#
#t_{flight} = " time of the entire flight of the bullet in seconds (s)"#

#"Note that the free fall under gravity is completely symmetric."#
#"This means that the time to reach the highest point is equal"#
#"to the time of falling from the highest point to the ground."#
#"This means that the time of the entire flight is the double of"#
#"the time to fall also."#