Question

A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fast is the diver moving while falling past a diving board that is 3.0 m above the surface of the pool?

Answer 1

The diver is heading downwards at 12 m/s Ignoring air resistance, the formula for the distance under constant acceleration is d = VT - 0.5AT^2 where V = initial velocity T = time A = acceleration (9.8 m/s^2 on Earth) In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m) So let's substitute the known values and solve for T d = VT - 0.5AT^2 -7 = 2.5T - 0.5*9.8T^2 -7 = 2.5T - 4.9T^2 0 = 2.5T - 4.9T^2 + 7 We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164. Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s V = 2.5 m/s - 14.47706141 m/s V = -11.97706141 m/s So the diver is going down at a velocity of 11.98 m/s Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point. V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s) V = 2.5 m/s - (-9.477061409 m/s) V = 2.5 m/s + 9.477061409 m/s V = 11.97706141 m/s And you get the exact same velocity, except it's the opposite sign. In any case, the result needs to be rounded to 2 significant figures which is -12 m/s