<h3>With a uniform acceleration of 2 m/s ²</h3>
Answer:
θ = 8.50°
To the nearest angle
θ = 9.0°
the golfer must hit the ball at angle 9° so that it travels 120 feet.
Explanation:
The range of a projectile is the horizontal distance covered by a projectile, which can be written as;
r = (u^2× sin2θ)/g
Where;
r = range
u = initial speed
θ = angle from horizontal
g = acceleration due to gravity
Solving for θ,
sin2θ = rg/u^2
θ = 1/2 × sin⁻¹(rg/u^2) ....1
Given;
r = 120 ft
u = 115 ft/s
g = 9.81m/s = 32.2 ft/s
Substituting the values into the equation 1;
θ = 1/2 × sin⁻¹(120×32.2/115^2)
θ = 1/2 × sin⁻¹(0.29217)
θ = 1/2 × 17.00
θ = 8.50°
To the nearest angle
θ = 9.0°
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Given:
f = 1160 kHz = 1160 x 10³ Hz
The velocity is c = 3 x 10⁸ m/s, the velocity of light (approximmately).
Calculate the wavelength, λ.
c = fλ
λ = c/f = (3 x 10⁸ m/s)/(1160 x 10³ 1/s) = 258.62 m
Answer: 258.6 m
