While traveling along a highway a driver slows from 32 m/s and comes to a stop with an acceleration of -6 m/s2. How long did it
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Answer:
The second ball lands 1.5 s after the first ball.
Explanation:
Given;
initial velocity of the ball, u = 12 m/s
height of fall, h = 35 m
initial velocity of the second, v = 12 m/s
Time taken for the first ball to land;
determine the maximum height reached by the second ball;
v² = u² -2gh
at maximum height, the final velocity, v = 0
0 = 12² - (2 x 9.8)h
19.6h = 144
h = 144 / 19.6
h = 7.35 m
time to reach this height;
Total height above the ground to be traveled by the second ball is given as;
= 7.35 m + 35m
= 42.35 m
Time taken for the second ball to fall from this height;
total time spent in air by the second ball;
T = t₁ + t₂
T = 1.23 s + 2.94 s
T = 4.17 s
Time taken for the second ball to land after the first ball is given by;
t = 4.17 s - 2.67 s
T = 1.5 s
Therefore, the second ball lands 1.5 s after the first ball.