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?
1 February, 17:49
39 cm / s
77.25 cm approx
Angle of projection θ = 47.7°
Maximum height H = 42.4 cm
Initial velocity = u = ?
we know that
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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» 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.