Explanation:
- Newton's first law of motion:
"An object at rest (or in uniform motion) remains at rest (or in uniform motion) unless acted upon an unbalanced force
In this situation, we can apply Newton's first law to the keys of the keyboard that are not hit by the fingers of the man. In fact, as no force act on the keys, they remain at rest.
- Newton's second law of motion:
"The acceleration experienced by an object is proportional to the net force exerted on the object; mathematically:

where F is the net force, m is the mass of the object, and a its acceleration"
In this case, we can apply Newton's second law to the keys of the keyboard that are hit by the man: in fact, as they are hit, they experience a downward force, and therefore they experience a downward acceleration.
"Newton's third law of motion:
"When an object A exerts a force on an object B (action force), then object B exerts an equal and opposite force on object A (reaction force)"
Here We can apply Newton's third law to the pair of objects finger-key: in fact, as the finger apply a force on the key (action force), then the key exerts a force back on the finger (reaction force), equal and opposite.
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.
Solved your another question same like this with scaling to Cm this time we go with metre(m)
Scale factor
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