A student skateboarding at a constant speed of 5m/s throws ball vertically into the air at 4m/s. She catches the ball as it returns to the ground. Let the ball be released from her hand at time t=0, the moment she passes the marker. A) What is the initial velocity of the ball? B) How long is the ball in the air (time of flight)? C) How far past the marker is the student when she catches the ball (Range)? (Assume she catches the ball at the same height she throws it) I'm sure that the initial velocity is 5i + 4j. And i can get to, a = -9.8j v = -9.8t j + c v(0) = c = ucos90 i + usin90 j v = ucos90 i + (usin90 - 9.8t) j (angle is 90 as thrown straight up, i hope) In this case Im not sure how to incorporate the initial velocity into it Im sure if i got A) i could get the rest. Hopefully someone can help.