Answer:
Explanation:
f = 
T = 120 N
L = 3.00 m
(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m
(wow that's massive for a "rope")
f = 
)
f = 
/6 = 0.527 Hz
This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.
A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N
I beleive that the wave with higher frequency will have shorter wavelength
As this is the projectile motion means the motion of object under constant acceleration, so we use all eq. of motion under constant acceleration.
Angle above horizontal=θ= 35°
Initial speed= v₁ = 25 m/s
Time of flight= t= 2.55 s
Now,
from eq. of motion
x=x₁ + (v₁*cosθ)(t)+1/2 *a*t²
As, there is no acceleration in horizontal direction so a=0 and also initial displacement x₁=0
x= 25*cos(35)*2.55
x=52.22 m
