Explanation:
First, find the velocity of the projectile needed to reach a height h when fired straight up.
Given:
Δy = h
v = 0
a = -g
Find: v₀
v² = v₀² + 2aΔy
(0)² = v₀² + 2(-g)(h)
v₀ = √(2gh)
Now find the height reached if the projectile is launched at a 45° angle.
Given:
v₀ = √(2gh) sin 45° = √(2gh) / √2 = √(gh)
v = 0
a = -g
Find: Δy
v² = v₀² + 2aΔy
(0)² = √(gh)² + 2(-g)Δy
2gΔy = gh
Δy = h/2
Answer:
Explanation:
Please check the picture and consider straight lines
Answer:
KE=2.3 x 10⁻¹⁹ J
Explanation:
Given that
λ = 544 nm
λ' = 485 nm
The kinetic energy KE given as
KE= E - Ф
Where


h= 6.626 x 10⁻³⁴
C=3 x 10⁸ m/s
Now by putting the values


KE=2.3 x 10⁻¹⁹ J
This is kinetic energy.