# Accelerated Linear Motion

#### mcintyre_ie

Hey
Im having some trouble with this Accelerated Linear Motion question. I dont have any clue where i should even start with part (B).
(A) A particle moving in a straight line with constant acceleration passes three points, p, q and r and has speeds u and 7u and p and r respecitively.
(I) Find its speed at q, the midpoint of pr, in terms o u
(II)Show that the time from p to q is twice that from q to r
(B) A juggler throws up six balls, one after the other at equal intervals of time t, each to a height of 3m. The first ball returns to his hand t seconds after the sixth was thrown up and is immediately thrown to the same height, and so on continually. You may assume that each ball moves vertically.
Find:
(i) the initial velocity of each ball
(ii) the time t
(iii) the heights of the other balls when any one reaches the jugglers hand.
Im just lost, dont have a clue where to start. My head just starts to go into overload when i even think about it. Please help!

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#### nautica

V(final) = V (original) + acceleration * time

or

Vf - Vo + at

You have only 3 points but 4 velocities, so I assume, your original or initial velocity would be u.

Have you tried this formula???

Nautica

#### mcintyre_ie

Yes, ive used the formulae before, im having trouble with part two, the juggler question.

#### HallsofIvy

Part two is relatively simple.

Each ball has acceleration -g= -9.81 m/s2. If we take "v" to be the initial velocity of each ball, then v(T)= v- 9.81T in m/s where T is measured in seconds since the ball was thrown.
In addition, the height of each ball after T seconds is given by
h(T)= vT- (9.81/2)T2.
The ball will continue upward until its velocity is 0: v- 9.81T= 0 so
T= v/9.81. We are told that the height at that time is 3 m:
v(v/9.81)- (9.81/2)(v/9.81)2= v2/(2*9.81)= 3 so
v2= 2*3*9.81 and v= &radi;(2*3*9.81).

You can find t in either of two ways: directly- The ball comes back to the person's had (assumed same height) exactly t seconds after the sixth ball is throne up and there is and interval of t seconds between each one: the first ball comes back to the person's had after 6t seconds: h(6t)= v(6t)- (9.81/2)(6t)2= 0. You know v from the previous problem. Solve for t.
Less directly: because of the symmetry of the motion, each ball attains it highest point exactly half way through its motion: 3t.
Since you calculated that time, T, in the first problem, 3t= T. Solve for t.

You know that when the first ball is again in the juggler's hand, time 6t has passed. Its height is0. The second ball is exactly "t" behind. Its height will be given by h(5t), the third ball's height will be h(4t), the fourth ball's height by h(3t), the fifth ball's height by h(2t), and the sixth ball's height by h(t). If the juggler throws the first ball up again the instant he catches it, each ball will have one of those 6 heights each time a ball is caught.

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