Rephrased Question : A ball is hit with a velocity of 40 m/s at an angle of #26^{\circ}# above the horizontal. A fielder who is located 110 m from where the ball is hit can reach up to 3m above the ground. If the ball was 120 cm above the ground when it was hit, how high above the fielders glove does the ball pass?
Projectile Motions :
Horizontal Component : # x-x_0=v_{o}\cos\theta_ot;#
Vertical Component: # y-y_o = v_o\sin\theta_ot - 1/2 g.t^2;#
#x_o = 0m; y_o = 120 cm = 1.2m; \theta_o=26^{\circ}; v_o=40#m/s.
With this choice the horizontal component gives, #x=v_{o}\cos\theta_ot => t = x/(v_o\cos\theta_o)#
Substituting this expression for '#t#' in the vertical component,
#y-y_o=\tan\theta_o x - g/(2\v_o^2\cos^2\theta_o).x^2#
Vertical position of the ball at a distance of #x=110 m# is :
#y=y_{o}+\tan\theta_o x-g/(v_o^2\cos^2\theta_o) x^2; x=110m#
Substituting for the values of #y_o=1.2m#, #x=110m#,
#\theta_o=26^{\circ}#, #v_o=40#m.s, we get the vertical position of the ball as #y=8.98m#.
Since the fielder can jump up to #3.0m# from the ground, the ball will go #8.98m - 3.0m=5.98m# above the fielder's glove.
