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?

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

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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