What net force would be needed to accelerate the box at 2.3 m/s^2
2 answers:
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Answer:
The horizontal component of her velocity is approximately 1.389 m/s
The vertical component of her velocity is approximately 7.878 m/s
Explanation:
The given question parameters are;
The initial velocity with which Margaret leaps, v = 8.0 m/s
The angle to the horizontal with which she jumps, θ = 80° to the horizontal
The horizontal component of her velocity, vₓ = v × cos(θ)
∴ vₓ = 8.0 × cos(80°) ≈ 1.389
The horizontal component of her velocity, vₓ ≈ 1.389 m/s
The vertical component of her velocity, = v × sin(θ)
∴ = 8.0 × sin(80°) ≈ 7.878
The vertical component of her velocity, ≈ 7.878 m/s.
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
From the question we are told
If the average velocity during the athlete's walk back to the starting line in Guided Example 2.5 is – 1.50 m/s,
Generally the equation Time spent is mathematically given as
T=\frac{d}{v}
Therefore
Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back
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