A froghopper is a small insect which can launch itself upwards with an accelerat

Question

A froghopper is a small insect which can launch itself upwards with an acceleration of about 400 times gravity (see National Geographic News, July 2010). In the first stage of its jump, while its feet are in contact with the ground, the froghopper accelerates to reach a top speed of 4.0 m/s. This is reached when its body is 2 mm above the ground. After losing contact with the ground, acceleration is then only due to gravity. Calculate the following after the froghopper loses contact with the ground (ignore air resistance): The time to reach a maximum height The maximum height reached The speed on hitting the ground

Explanation / Answer

distnace moved is S = 2 mm = 2*10^-3 m

maximum speed is V = 4 m/sec

acclearation due to gravity is g = 9.8 m/s^2

using Kinematic equation

V^2 -Vo^2 = 2*a*S

4^2 - Vo^2 = 2*9.8*2*10^-3

Vo = 3.99 m/sec

1) time taken to reach the maximum height is t1 = Vo/g = (3.990/9.81) = 0.406 sec

2) maximum height reached is Hmax = Vo^2/(2*g) = 3.99^2/(2*9.81) = 0.811 m

3) Speed on hitting Ground is V = sqrt(2*g*Hmax) = sqrt(2*9.81*0.811) = 3.988 m/sec

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