Practice Problems: Free Fall Solutions

1. A rock is dropped from a garage roof from rest. The roof is 6.0 m from the ground.
a. (easy) Determine how long it takes the rock to hit the ground.
y - y_o = v_ot + ½at²
-6 - 0 = 0 + ½(-9.8)t²
t = 1.1 s
b. (easy) Determine the velocity of the rock as it hits the ground.
v = v_o + at
v = 0 + (-9.8)(1.1)
v = -10.8 m/s
c. (moderate) A second rock is projected straight upward from ground level at the moment the first rock was released. This second rock had an initial upward velocity of +6.0 m/s. How long will this second rock take to reach maximum height?
v = v_o + at
At max height v = 0
0 = 6 + (-9.8)t
t = 0.61 s
d. (hard) At what time after release will the two rock have the same height?
y₁ - y_o,1 = v_o,1t + ½at²   AND  y₂ - y_0,2 = v_o,2 t + ½at²
When y₁ = y₂ ...  y_o,1 + v_o,1t + ½at² =  y_0,2 + v_o,2 t + ½at²
0 + 0 + (-4.9)t² = -6 + 6t + (-4.9)t²
6 = 6t
t = 1s

2. (moderate) A projectile is shot upward at 189 m/s from a height of 20 m off the ground. How long will it take for the projectile to be at ground level?
y - y_o = v_ot + ½at²
0 - 20 = 189t - 4.9t²
4.9t² - 189t - 20 = 0
Use the quadratic equation:
t = {38.7s, -0.1 s}
The root that makes physical sense is t = 38.7 s

3. (moderate) A hot-air balloon is hovering over a large public park. The operator of the balloon then makes the balloon begin to rise at a constant rate of 0.8 m/s. At some height from the ground the operator drops a rock from the basket. The rock take 10.3 seconds to hit a target on the ground. How high (above the ground) was the rock when it was released and what was its maximum height?
To find initial height:
y - y_o = v_ot + ½at²
0 - y_o = 0.8(10.3) - 4.9(10.3)²
y_o = 511.60 m
To find max height: v = 0
v² = v_o² + 2gΔy
0 = (0.8)² + 19.6(y_max - 511.60)
y_max = 511.63 m  (the rock moved 30 cm upward before falling downward)