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~~74.9m/s.
x~~74.9 m/s is the horizontal distance the ball travels.
