Explanation:
initial height, yo = 2 m
initial velocity, u = 20 m/s
angle of projection,θ = 5 degree
distance of net = 7 m
height of net = 1 m
Let it covers a vertical distance y in time t .
Use Second equation of motion for vertical motion
As it hits the ground in time t, so put y = 0
Taking positive sign, t = 0.84 s
The ball travels a horizontal distance x in time t
X = 20 Cos5 x t
X = 16.76 m
As this distance is more than the distance of net, so it clears the net.
Let t' be the time taken to travel a horizontal distance equal to the distance of net
7 = 20 cos5 x t'
t' = 0.35 s
Let the vertical distance traveled by the ball in time t' is y'.
So,
y' = 2.008 m
So, it clears the net which is 1 m high.
It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m
your welcome, and have a great day.
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by

where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is

And the frequency of the motion instead is given by:

(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:

where
F is the force applied initially to the spring, so it is equal to the weight of the block:

k = 160 N/m is the spring constant
Solving for A, we find

Answer:

Explanation:
Assuming there is no waste of energy:
