A 50-lb. ocean probe in the shape of a sphere of radius r = 1 ft. is dropped int

Question

A 50-lb. ocean probe in the shape of a sphere of radius r = 1 ft. is dropped into the ocean from an altitude of 20 feet. At t = 0 it enters the water and is subjected to the buoyant force B and a drag force D due to its motion through the fluid. Find its velocity 30 seconds after it enters the ocean. You are given the following information: 1. Since the probe is dropped from a relatively small altitude, assume it free-falls until it enters the water, that is, assume no air resistance and so its initial velocity v(0) at t = 0 is just the free fall velocity attained after falling 20 feet. To find v(0), you can use the formula v^2_f = v^2_0 + 2 as which gives us the final velocity of an object whose initial velocity is v_0 after it travels a distance s with constant acceleration a. From Archimedes principle, the buoyant force is equal in magnitude to the weight of water displaced. Ocean water has weight density of approximately 62.4 lb/ft^3 and this spherical probe has volume V = 4/3 pi. We assume a drag force of the form D = -1.5v G = The force of gravity, that is, it's just the weight of the probe. It will help to set a coordinate system whose +y axis points down, as shown in the figure.

Explanation / Answer

Mass of probe =50Ib

Distance= 20ft

Velocity attained by probe after travelling 20ft distance In air wil be v^2-u^2=2as

u= 0so v^2 =2×g×s=2×32.174×20=1286.96

V=35.87ft/s

Bouyant force = weight of the fluid displace by it =Mg==density of fluid×volume×g =62.4×4/3×3.14×(1)^3=26.5Ibft/s^2

As given in dig also net force =force due to gravity -drag force-bouyant force =mg-26.5-1.5×35.87=1528.4

Ma =1528.4

a=30.6ft/s^2

V=u+at

V=35.87 +30.6×30=952.9ft/s

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