A student on the balcony of his apartment building (height of 10 m) launches a tennis ball straight up at a velocity of 29.4 m/s. What is the ball's speed when it reaches the sidewalk below?

Astudent throws a water balloon with speed v0 from a height h = 1.76 m at an angle θ = 21° above the horizontal toward a target on the ground. the target is located a horizontal distance d = 9.5 m from the student’s feet. assume that the balloon moves without air resistance. use a cartesian coordinate system with the origin at the balloon's initial position. (a) what is the position vector, rtarge t, that originates from the balloon's original position and terminates at the target? put this in terms of h and d, and represent it as a vector using i and j. (b) in terms of the variables in the problem, determine the time, t, after the launch it takes the balloon to reach the target. your answer should not include h. (c) create an expression for the balloon's vertical position as a function of time, y(t), in terms of t, vo, g, and θ. (d) determine the magnitude of the balloon's initial velocity, v0, in meters per second, by eliminating t from the previous two expressions.