(1) The time of motion of the arrow is 0.25 s.
(2) The vertical height dropped by the arrow as it approaches the target is 0.31 m.
The given parameters:
- <em>Horizontal distance of the arrow, X = 20 m</em>
- <em>Horizontal speed of the arrow, v = 80 m/s</em>
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The time of motion of the arrow is calculated as follows;

The vertical height dropped by the arrow as it approaches the target is calculated as follows;

Learn more about time of motion of projectile here: brainly.com/question/1912408
The equation for this is very simple you add then you subtract then you get the answer then you divide then it all works out for you
Answer:
Velocity=14[m/s]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy.
In the attached image we can see the illustration of the ball falling from the height of 20 meters, at this time the potential energy will have the following value.
![Ep=m*g*h\\where:\\m=3[kg]\\h=20[m]\\](https://tex.z-dn.net/?f=Ep%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cm%3D3%5Bkg%5D%5C%5Ch%3D20%5Bm%5D%5C%5C)
![Ep=3*9.81*20\\Ep=588.6[J]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A20%5C%5CEp%3D588.6%5BJ%5D)
When the ball passes through half of the distance (10m) its potential energy will have decreased by half as shown below.
![Ep=3*9.81*10\\Ep=294.3[m]](https://tex.z-dn.net/?f=Ep%3D3%2A9.81%2A10%5C%5CEp%3D294.3%5Bm%5D)
If we know that potential energy is transformed into kinetic energy, we can find the value of speed.
![Ek=\frac{1}{2} *m*v^{2} \\therefore\\v=\sqrt{\frac{Ek*2}{m} } \\v=\sqrt{\frac{294.3*2}{3} } \\\\v=14[m/s]](https://tex.z-dn.net/?f=Ek%3D%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%20%5C%5Ctherefore%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7BEk%2A2%7D%7Bm%7D%20%7D%20%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B294.3%2A2%7D%7B3%7D%20%7D%20%5C%5C%5C%5Cv%3D14%5Bm%2Fs%5D)
(A) The total initial momentum of the system is
(1.30 kg) (27.0 m/s) + (23.0 kg) (0 m/s) = 35.1 kg•m/s
(B) Momentum is conserved, so that the total momentum of the system after the collision is
35.1 kg•m/s = (1.30 kg + 23.0 kg) <em>v</em>
where <em>v</em> is the speed of the combined blocks. Solving for <em>v</em> gives
<em>v</em> = (35.1 kg•m/s) / (24.3 kg) ≈ 1.44 m/s
(C) The kinetic energy of the system after the collision is
1/2 (1.30 kg + 23.0 kg) (1.44 m/s)² ≈ 25.4 J
and before the collision, it is
1/2 (1.30 kg) (27.0 m/s)² ≈ 474 J
so that the change in kinetic energy is
∆<em>K</em> = 25.4 J - 474 J ≈ -449 J
Answer:
The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is attached to a neutral atom or molecule in the gaseous state to form a negative ion.