Question

A ball is thrown upward from the ground. You observe the ball through a window on its way up, and notice that it was visible for 1 seconds while it travels from the bottom of the window to the top, which is a length of 17.805 metres.

(a) How much time does it take for the ball to be seen again, in seconds?

(b) How far above the top of the window will the ball reach, in metres?

Answer 1

for window - l = 17.805 m, t= is, a=-g= -9.8 m/s² from v ce? = 2a5 --2x9:8x17.805 = -348.978140 also V-UE at --9-8x1 = -98 - 0 9 4 V+60= -348.978. - 35.61 - Leo from egon o f@ v= 35 . 35.61-9.8 = 12.905 m/s - 0 = 12.905 +9.8 = 22.705 mis -9.8 @ Since U= 12.905 mls, a = -9.8m/s² time to see again = time of flight above to be 2x12.905 = 2.645 9.8 3 [T = 3.64] © | V=12.905 mks, W=0, 0 = -9.8m/s? 0- (12.405) = 2x(-18) » , 166.54 8 h = .49 m = 8.5m above top of Lindoro = 2x9.8