Julie throws a ball to her friend Sarah. The ball leaves Julie\'s hand a distanc
ID: 1951389 • Letter: J
Question
Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 10 m/s at an angle 64 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.1)What is the horizontal component of the ball’s velocity when it leaves Julie's hand?
2)What is the vertical component of the ball’s velocity when it leaves Julie's hand?
3)What is the maximum height the ball goes above the ground?
4)What is the distance between the two girls?m
5)After catching the ball, Sarah throws it back to Julie. The ball leaves Sarah's hand a distance 1.5 meters above the ground, and is moving with a speed of 7 m/s when it reaches a maximum height of 7 m above the ground.
What is the speed of the ball when it leaves Sarah's hand?
6)How high above the ground will the ball be when it gets to Julie? (note, the ball may go over Julie's head.)m
Explanation / Answer
Initial height h = 1.5 m Initial speed u = 10 m / s Angle = 64 degrees 1)The horizontal component of the ball’s velocity when it leaves Julie's hand = u cos = 4.38 m / s 2)The vertical component of the ball’s velocity when it leaves Julie's hand = u sin = 8.987 m / s 3)The maximum height the ball goes above the ground H = h + [ ( u sin ) 2 / 2g ] = 1.5 m + 4.121 m = 5.62 m4)The distance between the two girls R = range = u 2 sin 2 / g = 8.04 m 5)Speed at maximum height U = 7 m/s Maximum height h ' = 7 m Let the initial speed be v then ( v sin ' ) 2 / 2g = 7 v sin ' = [ 7 x 2g] = 11.71 m / s ------( 1) v cos ' = U v cos ' = 7 -------( 2) Eq ( 1) / eq( 2) ==> tan ' = 1.673 ' = 59.13 degrees The speed of the ball when it leaves Sarah's hand v = 7 / cos ' = 13.64 m / s The speed of the ball when it leaves Sarah's hand v = 7 / cos ' = 13.64 m / s
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