A Physics 111 student is riding on a flatcar of a train traveling along a straight horizontal track at a constant speed of 12 m/s. The student throws a ball into the air on a trajectory that she observes to make an initial angle of 53° with respect to the horizontal along the same line as the track. The student's TA, who is standing on an embankment nearby, observes the ball to rise straight up vertically. How high does the ball rise in meters? ----------------------- Of course, I tried this equation, (vy)^2=(v0 sin theta)^2-2(g)(y-y0) where vy is final velocity in y component v0=initial velocity g=9.8 m/sec/sec y0=0 So if the ball reaches top of its height, its velocity is zero, thus 0=(12*sin 53)^2-(2*9.8)(y) and I get y=4.69 meters. Okay, this isn't right, and I'm probably going at this problem the wrong way, right? Argh!