A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °
B. The arrow will go over the branch.
<h3>A. How to determine the angle</h3>
- Range (R) = 74 m
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = ?
R = u²Sine(2θ) / g
74 = 33² × Sine (2θ) / 9.8
Cross multiply
74 × 9.8 = 33² × Sine (2θ)
725.2 = 1098 × Sine (2θ)
Divide both sides by 1098
Sine (2θ) = 725.2 / 1098
Sine (2θ) = 0.6605
Take the inverse of sine
2θ = Sine⁻¹ 0.6605
2θ = 41.3
Divide both sides by 2
θ = 41.3 / 2
θ = 20.7 °
<h3>B. How to determine if the arrow will go over or under the branch</h3>
To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = 20.7 °
- Maximum height (H) = ?
H = u²Sine²θ / 2g
H = [33² × (Sine 20.7)²] / (2 ×9.8)
H = 6.94 m
Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).
Therefore, we can conclude that the arrow will go over the branch
Learn more about projectile motion:
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Answer:
Explanation:
Moment of inertia of given shell
where
M represent sphere mass
R -sphere radius
we know linear speed is given as 
translational 
rotational 
total kinetic energy will be
fraction of rotaional to total K.E
We getting addicted so it keeps us from getting bored
Answer:
I think he should move the purple car to the right. good luck
