Anyone who answers this correctly, i will mark as brainiest.
1 answer:
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Answer:
B) d = √ ( 4 h₂ ( H - 2h₂))
Explanation:
A) 1) If the height of the slide is very small, there is no speed to leave the table, therefore do not recreate almost any horizontal distance
2) If the height is very small downwards, it touches the earth a little and the horizon is small,
B) to find an equation for horizontal distance (d)
We must maximize the speed at the bottom of the slide let's use energy
Starting point Higher
Em₀ = U = m g h₁
Final point. Lower (slide bottom)
Emf = K + U = ½ m v² + m gh₂
As there is no friction the energy is conserved
mgh₁ = ½ m v² + mgh₂
v² = 2 g (h₁-h₂)
This is the speed with which the block leaves the table, bone is the horizontal speed (vₓ)
The distance traveled when leaving the table can be searched with kinematics, projectile launch
x = v₀ₓ t
y = t - ½ g t²
The height is the height of the table (y = h₂), as it comes out horizontally the vertical speed is zero
t = √ 2h₂ / g
We substitute in the other equation
d = √ (2g (h₁-h₂)) √ 2h₂ / g
d = √ (4 h₂ (h₁-h₂))
H = h₁ + h₂
h₁ = H -h₂
d = √ ( 4 h₂ ( H - 2h₂))