1.A baseball is thrown with an initial veocity 40 m/s that makes an ange 0f 60° above the horizontal. (Use: g = 10 m/s^2 , sin60° = 0.87, cos60°= 0.5).Find The vertical and horizontal components of ts velocity just after 2 seconds

Given:
Initial velocity (v₀) = 40 m/s
Launch angle (θ) = 60°
Acceleration due to gravity (g) = 10 m/s²
sin(60°) = 0.87
cos(60°) = 0.5
Time (t) = 2 seconds

To find the vertical component of velocity (vᵥ):
vᵥ = v₀ * sin(θ)
vᵥ = 40 m/s * 0.87
vᵥ = 34.8 m/s

To find the horizontal component of velocity (vₕ):
vₕ = v₀ * cos(θ)
vₕ = 40 m/s * 0.5
vₕ = 20 m/s

Therefore, just after 2 seconds, the vertical component of velocity is approximately 34.8 m/s and the horizontal component of velocity is 20 m/s.

1.A baseball is thrown with an initial veocity 40 m/s that makes an ange 0f 60° above the horizontal. (Use: g = 10 m/s^2 , sin60° = 0.87, cos60°= 0.5).Find The vertical and horizontal components of ts velocity just after 2 seconds​

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1.A baseball is thrown with an initial veocity 40 m/s that makes an ange 0f 60° above the horizontal. (Use: g = 10 m/s^2 , sin60° = 0.87, cos60°= 0.5).Find The vertical and horizontal components of ts velocity just after 2 seconds