Question

Problem 1: ball. ------- Chrişi + skateboard bug Chris is on a skateboard that is traveling at a constant horizontal speed y, = 4.35 m/s. They hold a ball in their hand at a height of h = 1.22 meters above the ground and then they toss the ball "up" (as seen in their frame) so that they catch it back in their hand exactly 2.85 seconds later as show above (figure not shown to scale!) a) Relative to Chris' hand, what is the speed of the ball as it it travels upward? Explain your answer. b) Relative to the fixed ground, what is the maximum height of the ball in the air? Explain your answer. c) Suppose there is a tiny motionless bug on the ground immediately below Chris' hand at the instant when they toss the ball upward. Suppose we define standard unit vectors i and ſ corresponding to the frame of reference for the bug. Write down vector expressions for the position, velocity, and acceleration of the ball (r, Ū, and a) as a function of time, t as seen by the bug at the instant of time just before the ball is caught by Chris. Explain your answer.

Note: I believe part C is referring to the the skateboard running over the bug and therefore the bug traveling with the skateboard.

Answer 1

Relative to come hand the ball will have only vertical component q speed because the horizontal speed of the came and all is same because the cars catches the ball again. ut instial wertical speed be uly - Vy = ly - g) » lyle vertiend speed at highest point 0 42LAG Octly - 9.81x2.45 | 4y= 14.0 m) Relative to ground maximum height is. hy he's lyt - gt² -122= 14x2.45 – x9.142.85) 114=112 m Ø Just before the ball is caught the bug is at * Position vector 8 = ocity au (4.8 5X2.85) m = 12.4m y = 1122 m. . - (12:4€ +122 g Jom Bing na Velouty D - Vock+ Vyf Intial polition 19= 1-22M P= (4.352 - 14.099 m/s Accelerament, ata ase ê tay 9 (o? - ggg me? la= -9.81 9 m/s2