Physics help please Thank you?

Thank you so much :)

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3 Answers

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  • 6 months ago
    Favorite Answer

    (a)

    The optimal angle is 45°

    (b)

    Given: v₀ = 50 m/s at 45°

    Find the time to maximum height using the initial velocity.

    vᵧ = v₀ᵧ – gt

    0 = 50sin(45°) - (9.8 m/s²)t

    t = 3.61s

    Time to maximum distance is twice the time to maximum height.

    s = v₀ₓt

    s = 50cos(45°)(3.61)(2)

    s = 255 m

  • 6 months ago

    Using general results the horizontal range is given by:

    (v₀²/g)sin2θ

    The maximum value is when sin2θ = 1

    i.e. 2θ = 90°, so when θ = 45°   

    Hence, maximum range is v₀²/g

    so, (50)²/g => 255.1 metres

    :)>

  • oubaas
    Lv 7
    6 months ago

    vertical level motion :

    0 = Vo*sin Θ *t -g/2*t^2 

    Vo*sin Θ = g/2*t 

    t = 2Vo*sin Θ / g 

    horizontal motion :

    range = Vo*cos Θ*t = Vo*cos Θ*(2Vo*sin Θ / g) = Vo^2/g*2*sin Θ*cos Θ 

    trigonometry helps us , since 2*sin Θ*cos Θ = sin 2Θ

    sin 2Θ is max and equal to 1 when 2Θ = 90° , which means Θ = 45°

    max range = 50^2/g = 2500/9.8066 ≅ 255 m 

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