We use -g to denote the acceleration due to the force of gravity. At time t = 0
ID: 2876624 • Letter: W
Question
We use -g to denote the acceleration due to the force of gravity. At time t = 0 a juggler throws a first ball into the air at a velocity of 3/2g. Find a function y_1 (t) which gives the height of the ball above its initial position at time t. At what time does the first ball return to the juggler's hands? At what time is the velocity of the first ball equal to zero? The juggler throws a second ball into the air which has position function y_2 (t) = -1/2g (t - 1)^2 + 3/2g (t - 1). At what time does the juggler throw the second ball? At what time do the two balls have equal height?Explanation / Answer
. Solution : 1. we use the the formula s = ut + at2 / 2 .Given u = 3g / 2 and acceleration a = - g
s = y1 (t) = (3g / 2 ) t - g t2 / 2
2 . the ball comes to his hans when s =0 => 3gt / 2- g t2 / 2 =0 => g t ( 3 / 2 - t / 2 ) = 0
=> t =0 or t= 3 secs . after 3 secs the ball will reach the persons hand
3 .and when t = 1 / 2 ( the time taken to reach he persons hand ) the ball will reach maximum height
when t = 3 / 2 secs where the velocity = 0
4 . when t = 1 y2(t) =0 thensecond ball will leave his hand when t = 1 secs
5 . let t be the time the 2 balls meet
ie y1(t) = y2(t) => 3gt / 2 - gt2 / 2 = 3g(t-1) / 2 - g (t-1)2 / 2 => 2t=4 or after 2 secs the balls meet
t = 2 secs after 2 secs the 2 balls will reach the same height