Ball-Catch A ball is thrown over the head of a person who...

Ball-Catch A ball is thrown over the head of a person who is standing a horizontal distance, d = 46.0 m, from the point where the ball was thrown. The initial ball's velocity is at an angle of ? = 60° with respect to the ground and has a magnitude of v0 = 25.0 m/s. As soon as the ball is thrown, the person runs with a time-varying acceleration whose component along the x-axis is given by ap(t) = B · t where B is a positive constant. The person catches the ball at exactly the same height it was thrown from. Assume that the air resistance is negligible and that the gravitational acceleration is directed downward and has magnitude g. 9. Find the time of flight for the ball (use the bold information to solve) 10. Find the Average velocity (magnitude and direction) after 1/4 of the flight time has elapsed From integrating the acceleration of the person two times, it can be shown Xp(t) = B · t^3 / 6 + d 11. Determine the value of the constant "B" EXTRA CREDIT SHOW THE INTEGRATION OF ap(t) = B · t to find the position of the person as function of time

Transcript

0:00 Hi.

00:02 Here in this given problem, first of all this is the horizontal ground over which a person is running with an accelerated motion towards right.

00:27 Here this is the point from where a ball is projected in the air with an initial velocity v -0 at an angle theta from the horizontal.

00:45 Gap of the point from where the ball is thrown at the instant the person starts running towards right.

00:57 This gap is d.

00:59 This gap is given as d is equal to 46 .0 meter angle theta that is 60 degree v0 is 25 .0 meter per second and acceleration in the motion of this person it varies as a function of time and this is a p t this is given as b into t where b is some constant here in the question number nine as the ball is caught at the same height by the person like this at this point as the ball is caught at the same height it was thrown from so we can find time of light which will be given by the expression 2 v0 into sine theta by g so plugging in all the known values this is two times of 25 .0 into sine 60 degree divided by which is 9 .8 and it comes out to be equal to 4 .42 second answer for the first part of the problem the time in which the person will catch the ball then in the second question question number 10 we have to find speed of the ball velocity of the ball along with its direction at a time t which is equal to one -fourth of its time of flight means this is 1 by 4 into 4 .42 approximately it will be 1 .1 second so it's horizontal component of the velocity of the ball that will remain constant as there is no acceleration along horizontal so v -o -x initial horizontal velocity that will be vfx, final horizontal velocity.

03:49 And this will be given as vo cos 60 degree, means 25 .0 into half.

03:58 Means this is 12 .5 meter per second.

04:03 Then we find initial vertical velocity, that will be vo.

04:10 Sin theta means 25 .0 into sine 60 degree.

04:17 And then using first equation of motion, final vertical velocity after a time 1 .1 second that will be given by using first equation of motion vfi is equal to v -o -y, initial vertical velocity minus g into t.

04:37 So this is v -o -sin theta minus g t or we can say this is 25 .0.

04:47 For sine 60 degree, this is root 3 by 2 minus 9 .8 and time, that is 1 .1.

04:56 So it comes out to be equal to 10 .9 meter per second.

05:02 Final vertical velocity after a time 1 .1 second.

05:07 So if these are the two components, this is v -o, this is vfx, final horizontal velocity, and this is final vertical velocity vfy so the net magnitude of the net velocity will be vf and at an angle alpha so this vf will be given by pythagorciorum vfx square plus vfy square means this is square root of square of 12 .5 plus square of 10 .9.

05:52 And here it will be 275 .06...

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