Can someone please help with a free fall question. "You throw a stone vertically upward with an initial speed of 6.0 m/s from the 3rd story office window. if the window is 12 m from the ground fine the time the stone is in flight and the speed of the stone just before it hits the ground"

Question

anonymous · Answer

Assuming there is no friction, linear kinematics tells us that $$y = y_o + v_ot + \frac{1}{2}g*t^2$$
Plugging in known information gives us $$0 = 12 + 6t + \frac{1}{2}(-9.81)t^2$$
Notice that this is a quadratic equation because it takes the form $$0 = ax^2 + bx + c$$
Therefore, you can solve for t as you would for x by using the quadratic formula:
$$x = \frac{-b\frac{+}{}\sqrt{b^2 - 4ac}}{2a}$$
For values a = -4.905, b = 6, and c = 12.

Solving for t tells us that the stone should contact the ground in 2.3 seconds.

anonymous · Answer

Ah, I missed the second part when reading: We need to find the velocity of the stone immediately before it contacts the ground.

Linear kinematics also tells us that $$v = v_o + gt$$
Since we know the stone's initial velocity (6m/s), the time of impact (2.3 seconds) and gravity (-9.81m/s^2), we can simply plug these in and solve for velocity:
$$v = v_o + gt = 6 +  (-9.81)2.3 = -16.6\frac{m}{s}$$
Notice that the impact velocity is negative. This is because the up direction is positive due to gravity being defined as negative (-9.81m/s^2).