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1 February, 16:46

When leaping at an angle of 47.7° above the horizontal, a froghopper reaches a maximum height of 42.4 cm above the level ground. What was the takeoff speed for such a leap?

What horizontal distance did the froghopper cover for this world-record leap?

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Answers (1)
  1. 1 February, 17:49
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    39 cm / s

    77.25 cm approx

    Explanation:

    Angle of projection θ = 47.7°

    Maximum height H = 42.4 cm

    Initial velocity = u = ?

    we know that

    maximum height

    H = U² x sin²θ / 2g

    U² = H x 2g / sin²θ

    Putting the values

    U² = (42.4 X 2 X9.8) / (sin47.7) ²

    U = 39 cm / s

    Horizontal Range R = U²sin2θ / 2g

    = 39 x 39 x (sin95.4) / 2 x 9.8

    R = 77.25 cm approx
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