Answer:
Explanation:
Let the velocity of firing be u at angle θ
At maximum height velocity will be equal to horizontal component of initial velocity or vcosθ
So , vtop = v cosθ
At height h/2
vertical component of velocity v₂
v₂² = (usinθ)² - 2 g . h/2
v₂² = u²sin²θ - gh
horizontal component of velocity at height h/2 = u cosθ
velocity at height h / 2
= √ ( u²sin²θ - gh + u² cos²θ)
Given
√ ( u²sin²θ - gh + u² cos²θ) = 2 vtop
u²sin²θ - gh + u² cos²θ = 4 v²top = 4 u² cos²θ
u²sin²θ - gh = 3 u² cos²θ
At height h , vertical component of velocity is zero
0 = u²sin²θ - 2gh
gh = u²sin²θ / 2
u²sin²θ - u²sin²θ / 2 = 3 u² cos²θ
u²sin²θ / 2 = 3 u² cos²θ
Tan²θ = 6
Tanθ = 2.45
θ = 68⁰ .
Answer:
Angle θ ≅ 21.5°
Explanation:
Given:
speed Vi= 60 m/s, Range R= 250 m, g=9.81 m/s²
To find:
Angle θ = ?
Sol: we know that Rang R = Vi² sin 2θ / g
⇒ Sin 2θ = g×R / Vi²
Sin 2θ = (9.81 m/s² × 250 m) / ( 60 m/s)²
Sin 2θ = 0.68125
2θ = Sin ⁻¹ (0.68125)
2θ = 42.9413
θ = 42.9413 / 2
θ = 21.4706 ≅ 21.5°
<span>The bending of light rays by sun is twice in Einstein's theory as in Newton's theory of gravitation. Einstein's theory was verified by Eddington during the solar eclipse in 1919
Also observed precession of aphelion of mercury completely explained by Einstein's theory but not by Newtonian gravity</span>