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11 February, 10:43

energy conservation-problems 1. A slingshot fires a pebble from the top of a building at a speed of 14.0m/is. The building is 31.0m tall. Ignoring all frictional effects, find the speed with which the pebble strikes the ground

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  1. 11 February, 12:15
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    Horizontal velocity is 14 m/s

    Vertical velocity is 28.3 m/s

    Explanation:

    Hello dear friend, you have not mentioned the type of speed that you require for this problem.

    In this case, there are three possibilities with which the slingshot can be fired i. e. Horizontally, Vertically straight up and vertically straight down. Below is the explanation / answer to all three possibilities

    Fired horizontally:

    initial conditions:

    Vertical Velocity = 0; Horizontal Velocity = 14m/s

    final conditions:

    Vertical Velocity (v² = u² + 2gs) but initial vertical velocity is zero

    v² = 2gs so v² = 2 (9.8) (31) = 607

    v = 24.6m/s

    but Horizontal Velocity is still = 14m/s

    Resultant velocity from these two velocity components (Pythagoras theorem)

    V² = (v horizontal) ² + (v vertical) ² = 14² + 24.6²

    V = 28.3m/s

    angle = tan ⁻¹ (24.3/14) = 60.1⁰

    V = 28.3m/s at angle of 60.1⁰ to the horizontal

    Fired Vertically Straight Up

    distance before the pebble reaches maximum height from top of building

    v² = u² + 2gs

    where, v is zero at maximum height

    g is minus for upward motion.

    v² = u² + 2gs

    0 = 14² - 2 (9.8) s

    s = 196/19.6 = 10.0m

    totals distance from maximum height to the ground = 10.0 m + 31.0 m = 41.0m

    v² = u² + 2gs

    now u from maximum height is 0 and g is positive for downward motion

    v² = 2gs

    v² = 2 (9.8) (41.0)

    v = 28.3m/s

    v = 28.3m/s vertically straight up

    Fired Vertically Straight Down

    v² = u² + 2gs

    u = 14m/s, g = 9.8m/s², s = 31.0m

    v² = 14² + 2 (9.8) (31.0)

    v = 28.3m/s

    v = 28.3m/s vertically straight down
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