Julie throws a ball to her friend sarah the ball leaves...

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 20 m/s at an angle 52 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 right before Sarah catches it? 2) What is the vertical component of the ball’s velocity right before Sarah catches it? 3)What is the time the ball is in the air? 4)What is the distance between the two girls? 5)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 17 m above the ground, and takes 2.826 s to get directly over Julie's head. 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?

Transcript

00:01 Hello, the question is taken from physics and the question is it contains various spot.

00:07 The ball leaves julie's hand at it distance 1 .5 meter above the ground with the initial speed of 12 meter per second at an angle 47 degree.

00:17 Okay, what is the horizontal component of ball's velocity when it just till the julie's end? so horizontal during the projected motion horizontal velocity always remain constant.

00:30 Okay, so horizontal velocity is vx is ux that is equal to u course of theta okay so that will be 20 course 52 degree which will gives us the value of horizontal velocity is 20 course 52 degrees so that is 12 .31 meter per second next what is the value of the vertical velocity just before okay so the situation let me first draw the situation so so it is at a height of 1 .5 meter.

01:18 So this is horizontal velocity, vertical velocity is okay so that is corresponding to u sine theta into t.

01:26 Okay, so but t we don't know so we are going to use the s is equal to a v square minus u square divided by two way so final velocity is when we need to evaluate the value of v.

01:42 So v square is equal to u square plus 2 as we already know and a we already know so we will be equal to square out u is the value of initial velocities sorry that is vertical component so this will be u y okay u y is 20 sine 52 degree plus 2 into a is 9 .8 as is 1 .5 so which will give us the value of the vertical velocity just before it hit the hand is so 20 sign 52 degree plus 2 into 9 .8 into 1 .5 so taking the square root of it so that is 6 .7 meter per second okay 7.

02:35 Okay so these are the required horizontal and vertical velocity so this is just wrong this is we can write it as we okay so third part is what is the time the ball is in the air now we need to evaluate the time of light okay so there are two time in this case so first is when it reaches at maximum height and second time it reaches from the maximum height to reach the hand of the gen so third part is first case is time to reach the maximum height is t is equal to first is vertical velocity that is once again we can evaluate by horizontal velocity also because horizontal acceleration is equal to zero but we didn't know that at maximum height vertical velocity is equal to zero so u is equal to g t u y so time to reach the maximum height is let us take this is t1 that is u y over g u is 20 sine 52 degree divided by 9 .8 this is the time to reach maximum height so that will be 20 sine 52 degree by 9 .8 so that is 1 .6 1 second now what is the time to reach in the hand of the chain on the maximum height so that is true so at maximum height what vertical velocity is equal to zero so time to reach from the maximum height to the position so first we need to evaluate the maximum height also so maximum height is u square sine square theta over g or u by 400 sine square 52 degree divided by 9 .8 so let me evaluate this value 25 .35 meter which is the maximum height okay now time to reach from this height to the 1 .5 meter is so that is as is equal to so we which formula we can use so v square minus u square but we need to evaluate the time v is equal to u plus 80 but we know the value of h so v square minus so let us first we square minus u square so s is equal to we will use h so that is delta h is equal to half g t square so because initial velocity is equal to zero that is the velocity at maximum high delta h is equal to 25 .35 minus 1 .5 multiply 2 divided by 9 .8 is equal to t 2 taking the square root of it.

05:58 So from here the value of t 2 will be 0 .35 minus 1 .5 so multiply with square root 2 divided by so that is 2 .2...

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