First, calculate how long the ball is in midair. This will depend only on the vertical displacement; once the ball hits the ground, projectile motion is over. Since the ball is thrown horizontally, it originally has no vertical speed.
t = time vi = initial vertical speed = 0m/s g = gravity = -9.8m/s^2 y = vertical displacement = -45m
y = .5gt^2 [Basically, in this equation we see how long it takes the ball to fall 45m] -45m = .5 (-9.8m/s^2) * t^2 t = 3.03 s
Now we know that the ball is midair for 3.03s. Since horizontal speed is constant we can simply use:
x = horizontal displacement v = horizontal speed = 25m/s t = time = 3.03s
x = v*t x = 25m/s * 3.03s = 75.76 m Thus, the ball goes about 75 or 76 m from the base of the cliff.
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Answer:
16.7 s
Explanation:
T= <u>Vf - Vo</u> a= <u>F</u>
a m
4,500 / 3000 = 1.5 (a)
30 - 5 / 1.5(a) = 16.7 s
Answer:
a) and c).
Explanation:
For a complete destructive interference occur, it must be met the following condition relating the wavelength, and the difference in the paths taken by the sound emitted by the sources until arriving to the listening point:
d = |dA- dB| = (2n-1)*(λ/2)
For n= 1, d = λ/2 = 0.25 m, it doesn't meet any of the cases.
For n=2, d= 3*(λ/2) = 0.75 m
In the case a) we have dA = 2.15 m and dB = 3.00 m, so dB-dA = 0.75 m, which means that in the location stated by case a) a complete destructive interference would occur.
For n=3, d= 5*(λ/2) = 5*0.25 m = 1.25 m.
This is just the case c) because we have dA = 3.75 m and dB = 2.50 m, so dA-dB = 1.25 m, which means that in the location stated by case c) a complete destructive interference would occur also.
The remaining cases don't meet the condition stated above, so the statements found to be true are a) and c),
