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:
brainly.com/question/20326485
#SPJ1
Answer:
Explanation:
The average kinetic energy of molecules in a pot increases when it is heated on a stove . The heat energy is spent to decrease the intermolecular attractive force . The result is increase in kinetic energy due to increase in the velocity of molecules
formula for kinetic energy of one mole of a monoatomic gas
= 3/2 RT
when temperature increases , kinetic energy increases .
This indicates nucleus of an atom constantly positively charged. an atom has an nuteral overall charge because it has the same amount of electrons as protons
They are called magnets poles.
Answer:
ΔP = 20000 N s
Explanation:
To solve this problem we use the relation between momentum and moment
I = Δp
let's calculate the momentum
I = ∫F dt
if we use the average force
I = F t
I = 10000 2
I = 20000 N s
therefore with the first equation
ΔP = I = 20000 N s