Answer:
the shooting angle ia 18.4º
Explanation:
For resolution of this exercise we use projectile launch expressions, let's see the scope
R = Vo² sin (2θ) / g
sin 2θ = g R / Vo²
sin 2θ = 9.8 75/35²
2θ = sin⁻¹ (0.6)
θ = 18.4º
To know how for the arrow the tree branch we calculate the height of the arrow at this point
X2 = 75/2 = 37.5 m
We calculate the time to reach this point since the speed is constant on the X axis
X = Vox t
t2 = X2 / Vox = X2 / (Vo cosθ)
t2 = 37.5 / (35 cos 18.4)
t2 = 1.13 s
With this time we calculate the height at this point
Y = Voy t - ½ g t²
Y = 35 sin 18.4 1.13 - ½ 9.8 1,13²
Y = 6.23 m
With the height of the branch is 3.5 m and the arrow passes to 6.23, it passes over the branch
F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps
Answer: The answer is A for sure
Explanation:
That is, there will be no acceleration. If you are sitting at rest in a chair and the upward push of the chair is equal to the downward pull of gravity, you will stay at rest in the chair. ... You now have an unbalanced force acting on you and therefore, according to Newton's First Law, your motion is going to change.
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Answer:
B. Mechanical energy= 50J+30J=80J
Answer:
Speed at which it will reach the ground is given as
Total time for which it will remain in air is given as
t = 6.3 s
Explanation:
As we know that the object is projected upwards with speed
now when it will reach the ground then we have
so we have
so we have
Now speed of the object when it reaches the ground is given as
