Respuesta :
The trajectory of a projectile is the path of a parabola which is bilaterally symmetrical
The initial speed of the grasshopper is approximately 1.566 m/s
The height of the cliff is approximately 4.43 m
The reasons the above values are correct are presented as follows:
The given parameters are;
The height of the leap of the grasshopper above the cliff, d₁ = 6.90 cm
The range of the leap of the grasshopper, d₂ = 1.12 m
The angle of the grasshopper's leap, θ = 48°
Required:
The initial speed of the grasshopper
Solution:
The initial speed is given b the equation, v² = (u·sin(θ))² - 2·g·h
Where;
v = The final velocity of the grasshopper = 0 at maximum height
u = The initial speed of the grasshopper
g = The acceleration due to gravity ≈ 9.81 m/s²
h = The maximum height reached from the cliff top= d₁ = 6.90 cm = 0.069 m
Therefore;
0² = (u×sin(48°))² - 2 × 9.81 × 0.069
(u×sin(48°))² = 2 × 9.81 × 0.069
u = (√(2 × 9.81 × 0.069))/(sin(48°)) ≈ 1.566
The initial speed of the grasshopper, u ≈ 1.566 m/s
Required:
The height of the cliff;
Solution:
The equation of projectile's trajectory is given as follows;
y = y₀ + u×sin(48°)×t - (1/2)·g·t²
The time it takes the grasshopper to cover the range, d₂, t = d₂/vₓ
Where;
vₓ = The horizontal velocity = u·cos(θ)
∴ t = 1.12/(1.566 × cos(48°) ≈ 1.0688
y₀ = The height of the cliff
0.069 = y₀ + 1.566×sin(48°)×1.0688 - (1/2)×9.81×1.0688²
y₀ = 0.069 - (1.566×sin(48°)×1.0688 - (1/2)×9.81×1.0688²) ≈ 4.43
The height of the cliff, y₀ ≈ 4.43 m
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