A projectile is fired vertically from earth's surface with an initial speed of 5.3 km/s. neglecting air drag, how far above the
1 answer:
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Answer:
A) 1.122 m
B) 5.4 hertz
C) 6.06 m/s ( towards negative x axis )
D) 0.22
E) max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0
Explanation:
The travelling wave = D = 0.22sin ( 5.6x + 34t)
this wave is represented in the form : D = Asin ( kx + wt)
A) wavelength
k = 2 / v
v = wavelength
k = 5.6
therefore wavelength ( v ) from the equation = 2 / 5.6 = 1.122 m
B ) frequency
w = 2f
w = 34
f = frequency
therefore f = w / 2 = 34 / 2
= 5.4 hertz
C ) velocity
speed = f v
v = 5.4 * 1.122 = 6.06 m/s ( towards negative x axis )
D) amplitude
A = 0.22
E ) maximum and minimum speeds of particles on the chord
max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0