A bird carrying a fish (5kg) drops it from 107 meters in the air how fast does the fish hit the ground
1 answer:
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<h2>Answer:</h2>
<em><u>Velocity of throwing arrow = 43.13 m/s.</u></em>
<h2>Explanation:</h2>In the question,
Let us say the height from which the arrow was shot = h
Distance traveled by the arrow in horizontal = 61 m
Angle made by the arrow with the ground = 2°
So,
From the <u>equations of the motion</u>,
Now,
Also,
Finally, the angle made is 2 degrees with the horizontal.
So,
Final horizontal velocity = v.cos20°
Final vertical velocity = v.sin20°
Now,
u = v.cos20° (No acceleration in horizontal)
Also,
So,
We can say that,
<em><u>Therefore, the velocity with which the arrow was shot by the archer is 43.13 m/s.</u></em>
Answer:
a. S(t)=350−1t
Explanation:
To determine the equation of motion you take into account the general form of motion with constant velocity:
( 1 )
So is the initial position from a specific reference frame. In this case is 350 m.
v is the speed of the motion, in this case is 1m/s. However, the motion is forward the zero point of the reference frame, hence, the speed is - 1m/s.
You replace the values of So and v in the equation ( 1 ) and you obtain:
Hence, the answer is:
a. S(t)=350−1t
- - - - - - - - - - - - - - - - - - - - -
Para determinar a equação do movimento, você leva em consideração a forma geral do movimento com velocidade constante:
(1)
Assim é a posição inicial de um quadro de referência específico. Neste caso, é de 350 m.
v é a velocidade do movimento, neste caso é de 1m / s. No entanto, o movimento é avançar o ponto zero do quadro de referência, portanto, a velocidade é de - 1m / s.
Você substitui os valores de So ev na equação (1) e obtém:
Portanto, a resposta é:
uma. S (t) = 350-1t, movimento retrógrado