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!
It looks to me like you are assuming vo= 12 m/s and there is no reason to do that. What you are told is that the ball rises vertically relative to the ground. Since the train is moving at 12 m/s, the HORIZONTAL COMPONENT of the ball (relative to the train) must be 12 m/s. Use that to find vo. (The student is, by the way, throwing the ball opposite to the direction of the train.)