What is a smash shot? (BADMINTON)

1 answer:

5 0

The smash shot in badminton is considered one of the most powerful kind of shot that can tilt the score in your favour. This shot can only be excited when the shuttle is high in the air. The reason behind that is because from a high elevation point, the shuttle is smashed downward over the net into the opponent's court. There is almost no defence against such a shot because it is slammed so quickly and is at such a downward angle that it is difficult for the opponent to receive it.

You might be interested in

A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure

klasskru [66]

Answer:

$3x_0$

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

$d=v_x t$

where

$v_x$ is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

$d=v_0 t = x_0$

in the second case, the horizontal velocity is increased to

$v_x' = 3v_0$

And so the new distance travelled will be

$d' = v_x' t = 3 v_0 t = 3 x_0$

So, the distance increases linearly with the horizontal velocity.

5 0

4 years ago

When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink

Dovator [93]

Answer:

y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

Therefore the diffraction equation for slits with m = 1 remains

a sin θ = λ

in general these experiments occur for oblique angles so

sin θ = θ

θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

θ = s / R

as the angle is small the arc approaches the distance s = y

y / R = 1.22 λ / s

y = 1.22 λ R / a