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Sagot :
Answer:
maximum height = 5.1 m distance traveled/horizontal displacement = 35.3m
Explanation:
Vy = sin 30º • 20 m/s
Vy = 10 m/s
Vx = cos 30º • 20m/s
Vx = 17.3
V final = V initial + acceleration • time
0 = 10m/s + (-9.81 m/s/s) • time
subtract 10 m/s from both sides
-10m/s = -9.81m/s/s • time
divide each side by -9.81m/s/s
t = 10/9.81 s or roughly 1.02 second
the time to reach max height is equal to 1.02 s
Height = 1/2 • acceleration • time^2 + Vy • time
Height = 1/2 • (-9.81 m/s/s) • 1.02s^2 + 10m/s • 1.02 s
Height = 5.1 m
the maximum height of the ball is equal to 5.1m
since the ball starts from an initial height of zero and has a final height of 0 once it reaches the ground, the time needed to reach max height is the same amount of time to fall back down to earth. thus, time total is equal to 1.02s multiplied by 2
t total = 1.02s • 2
t total = 2.04s
horizontal displacement = Vx • time total
horizontal displacement = 17.3m/s • 2.04 s
Horizontal displacement = 35.3m
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