A person jumps vertically from rest off a diving board that is 5 m above the surface of a deep pool The person's speed reaches zero 2 in below the surface of the pool Find the magnitude of the acceleration of the person while in the water, assuming that it remains cominam!

To find the magnitude of the acceleration of the person while in the water, we can use the kinematic equation:<br /><br />v^2 = u^2 + 2as<br /><br />where:<br />v = final velocity (0 m/s, since the speed reaches zero)<br />u = initial velocity (unknown)<br />a = acceleration (what we need to find)<br />s = displacement (2 m, since the speed reaches zero 2 m below the surface of the pool)<br /><br />Since the person's speed reaches zero, we can use the equation:<br /><br />v^2 = u^2 + 2as<br /><br />0 = u^2 + 2a(2)<br /><br />0 = u^2 + 4a<br /><br />We need to find the initial velocity (u) to solve for the acceleration (a). We can use the equation:<br /><br />v^2 = u^2 + 2as<br /><br />0 = u^2 + 4a<br /><br />Since the person jumps vertically from rest off a diving board that is 5 m above the surface of a deep pool, we can use the equation:<br /><br />v^2 = u^2 + 2as<br /><br />0 = u^2 + 2a(5)<br /><br />0 = u^2 + 10a<br /><br />We have two equations:<br /><br />0 = u^2 + 4a<br />0 = u^2 + 10a<br /><br />We can solve these two equations simultaneously to find the acceleration (a).<br /><br />Subtracting the first equation from the second equation, we get:<br /><br />0 - 0 = (u^2 + 10a) - (u^2 + 4a)<br /><br />0 = 6a<br /><br />a = 0 m/s^2<br /><br />Therefore, the magnitude of the acceleration of the person while in the water is 0 m/s^2.