Jami asked in Science & MathematicsPhysics · 2 months ago

PHYSICS HELP please :(?

A person kicks a rock off a cliff horizontally with a speed of 20 m/s. It takes 7.0 seconds to hit the ground. Find:

a) The height of the cliff.

b) The final velocity of the rock just before it hits the ground.

c) The distance the rock travels horizontally before it hits the ground.

 

it needs to be drawn but I have no clue how. 

3 Answers

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  • fooks
    Lv 7
    2 months ago
    Favorite Answer

    Refer to the figure below.

    Take g = 9.8 m/s²

    Take all downward quantities to be positive.

    (a)

    Consider the vertical motion (uniform acceleration motion):

    Initial velocity, u(y) = 0 m/s

    Acceleration, a(y) = 9.8 m/s²

    Time taken, t = 7.0 s

    s(y) = u(y) t + (1/2) a t²

    s(y) = (0) (7) + (1/2) (9.8) (7)²

    Height of the cliff, s(y) = 240 m

    (b)

    Consider the vertical motion:

    v(y) = u(y) + a(y) t

    v(y) = (0) + (9.8) (7)

    v(y) = 68.6 m/s

    Final velocity of the rock

    = √[v(x)² + v(y)²]

    = √[20² + 68.6²]

    = 71.5 m/s

    Make an angle of tan⁻¹(68.6/20) = 73.7° below the horizontal line.

    (c)

    Consider the horizontal motion (uniform velocity motion):

    Distance the rock travels horizontal before it hits the ground (range)

    = v(x) t

    = (20) (7)

    = 140 m

    Attachment image
  • 2 months ago

    a) The height of the cliff.

    g = gravitational acceleration = 9.81 m/s²

    viy = initial velocity in the vertical direction = 0 m/s

    t = elapsed time of fall = 7.0 s

    Δy = height of cliff (vertical distance) = to be determined

    Δy = viy∙t + 0.5gt²

    Δy = 0 m/s∙7.0 s + 0.5(9.81 m/s²)(7.0 s)²

    Δy = 240.345 m = 240 m

    b) The final velocity of the rock just before it hits the ground.

    vfx = velocity in the horizontal direction = 20 m/s

    vfy = final velocity in the vertical direction = to be determined

    Vr = resultant final velocity = to be determined

    vfy = viy + gt

    vfy = 0 m/s + (9.81 m/s²)(7.0 s)

    vfy = 68.67 m/s = 69 m/s

    Vr = √[(vfx)² + (vfy)²]

    Vr = √[(20 m/s)² + (69 m/s)²]

    Vr = 71.52320533 m/s = 72 m/s

    c) The distance the rock travels horizontally before it hits the ground.

    vfx = velocity in the horizontal direction = 20 m/s

    t = elapsed time of fall = 7.0 s

    Δx = horizontal distance = to be determined

    vfx = Δx / t

    vfx∙t = Δx

    Δx = vfx∙t

    Δx = 20 m/s∙7.0 s

    Δx = 140 m

  • Jim
    Lv 7
    2 months ago

    y(t) = ½at² + v₀t  + y₀ is the basic formula you need to know.

    Standard gravity is -9.80665 m/s²

    For speed KE=PE, ie 1/2mv² = mgh

    Here are the resolved Kinetics formulas:

    Attachment image
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