Explanation:
y = y₀ + v₀ t + ½ at²
For the first ball:
0 = h + v₀ t − 4.9t²
For the second ball:
0 = h − 4.9 (t−1)²
a) If h = 20.0, find v₀.
0 = 20 − 4.9 (t−1)²
t = 3.02 s
0 = 20 + v₀ (3.02) − 4.9 (3.02)²
v₀ = 8.18 m/s
Graph:
desmos.com/calculator/uk1wzkxybt
If v₀ is given, find h.
First, find t in terms of v₀:
h + v₀ t − 4.9t² = h − 4.9 (t−1)²
v₀ t − 4.9t² = -4.9 (t−1)²
v₀ t − 4.9t² = -4.9 (t² − 2t + 1)
v₀ t − 4.9t² = -4.9t² + 9.8t − 4.9
v₀ t = 9.8t − 4.9
(9.8 − v₀) t = 4.9
t = 4.9 / (9.8 − v₀)
Therefore:
h = 4.9 (4.9 / (9.8 − v₀) − 1)²
bi) If v₀ = 6.0 m/s:
h = 4.9 (1 / (9.8 − 6.0) − 1)²
h = 2.66 m
bii) If v₀ = 9.5 m/s:
h = 4.9 (1 / (9.8 − 9.5) − 1)²
h = 26.7 m
c) As found in part a, the time it takes for the first ball to land is:
t = 4.9 / (9.8 − v₀)
If v₀ is greater than 9.8 m/s, the time becomes negative, which isn't possible. Therefore, vmax = 9.8 m/s. At this speed, the ball would reach its highest point after 1 second, the same time that the second ball is dropped. Two balls dropped at the same time from different heights cannot land at the same time.
d) If v₀ is less than 4.9 m/s, the time for the first ball to land becomes less than 1 second. Which means it will have already landed before the second ball is dropped. Therefore, vmin = 4.9 m/s.
