A student in the lab has a 17 kg mass attached to a horizontal spring on a frictionless track as shown. The position labeled "O m" represents the unstrained position of the spring. The student initially moves the block so that it compresses the spring,
where the initial position is shown. The student releases the block from rest and starts a timer. The block moves to the
maximum position shown before returning to its initial position. The student counts the number of times that the block
returns to its initial position and finds that it takes 25.9 s for the block to return 14 times.
-3.00 m
0 m
+3.00 m
What is the magnitude of the maximum acceleration for this object in m/s2)?

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

i. like it i'm gonna edit it tomorrow

gravity

explanation:

their crystal patterns have been disrupted by stress