Question

Locusts can jump to heights of 0.870 m.

A) Assuming the locust jumps straight up, and ignoring air resistance, what is the takeoff speed of the locust?

b) The locust actually jumps at an angle of about 55.0° to the horizontal, and air resistance is not negligible. The result is that the takeoff speed is about 40.0% higher than the takeoff speed when the locust jumps straight up, and ignoring air resistance. If the mass of the locust is 2.00 g and its body moves 4.00 cm in a straight line while accelerating from rest to the takeoff speed, calculate the acceleration of the locust (assumed constant).

c)The locust actually jumps at an angle of about 55.0° to the horizontal, and air resistance is not negligible. The result is that the takeoff speed is about 40.0% higher than the takeoff speed when the locust jumps straight up, and ignoring air resistance. If the mass of the locust is 2.00 g and its body moves 4.00 cm in a straight line while accelerating from rest to the takeoff speed, estimate the force exerted on the hind legs by the ground (Ignore the locust’s weight).

Answer 1

a)

Jump Height = 0.870 m

PE = 0.870 m x M x 9.8 m/s^2

KE = 0.5 x M x V^2

V = sqrt(0.870 m x M x 9.8 m/s^2/(0.5 x M ) =. 4.13 m/s

b)

angle of jump = 55.0 deg.

Take of speed = 1.4 V = 4.13 m/s x 1.4 = 5.78 m/s

Distance moved in straight line to acquire take off speed of 5.78 m/s = 4.00 cm = 0.004 m

Acceleration of locust = (5.78^2-0^2)m^2/s^2 / ( 2 x 0.004m) = 4176 m/s^2

c)

Force reuired to be impressed on locust by ground = Force Locust has to apply on ground = 0.002Kg x 4176 m/s^2

=8.35 N