The equation that describes the motion of a projectile near the surface of the earth neglecting air resistance is
#y=y_0+xtantheta-(gx^2)/(2v^2cos^2theta)#
where
#y# = the vertical distance or the projectile from the ground.
#y_0# = the height from which the projectile was launched = 1 m in this problem.
#x# = the horizontal position of the projectile.
#theta# = the angle that the direction of the projectile's initial velocity makes with the ground = #36.9^@# in this problem.
#g# = the gravitational acceleration at the surface of the earth #~~ 9.8m/s^2#.
#v# = the initial magnitude of the velocity of the projectile = 27.4 m/s in this problem.
We want to know the #x# value when #y=0#. This means we need to set #y=0# and then solve for #x#.
#0=y_0+xtantheta-(gx^2)/(2v^2cos^2theta)#
Use the quadratic formula to solve for #x#.
#x=(-tanthetapmsqrt(tan^2theta-4(-g/(2v^2cos^2theta))y_0))/(-2(g)/(2v^2cos^2theta))#
#=(-tanthetapmsqrt(tan^2theta+(2gy_0)/(v^2cos^2theta)))/(-g/(v^2cos^2theta))#
#=(-tan(36.9^@)pmsqrt(tan^(2)(36.9^@)+(2*9.8*1)/((27.4)^2cos^(2)(36.9^@))))/(-9.8/((27.4)^2cos^(2)(36.9^@)))#
The two values we calculate are #x~~-1.3# and #x~~74.9#m/s.
#x~~74.9# m/s is the horizontal distance the ball travels.
