The dart player throws a dart horizontally at a speed of 24.75 m/s.
In this question,
The horizontal distance traveled by the dart, x = 2.5 m
The dart hits the board 5 cm below the height from which it was thrown.
⇒ 5 cm = 0.05 m
Let the time taken by the dart to hit the board is t.
The time taken can be calculated using the Newton's second law of motion,
[tex]y = ut+\frac{1}{2}at^{2}[/tex]
Put, u = 0, a = -g and g = 9.8
On substituting the above values, we get
⇒ [tex]-0.05 = 0-\frac{1}{2}(9.8)t^{2}[/tex]
⇒ [tex]0.05(\frac{2}{9.8}) = t^{2}[/tex]
⇒ [tex]t^{2} =0.0102[/tex]
⇒ [tex]t=\sqrt{0.0102}[/tex]
⇒ t = 0.1010 s
Now, the velocity (speed) can be calculated as
displacement = speed × time
⇒ 2.5 = speed × 0.1010
⇒ speed = [tex]\frac{2.5}{0.1010}[/tex]
⇒ speed = 24.7524 m/s
Hence we can conclude that the dart player throws a dart horizontally at a speed of 24.75 m/s.
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