An object whose mass is 100 lb falls freely under the influence of gravity from an initial elevation of 600 ft above the surface of earth. the initial velocity is downward with a magnitude of 50 ft/s. the effect of air resistance is negligible. determine the velocity, in ft/s, of the object just before it strikes earth. assume g = 31.5 ft/s

Answer with Explanations:

Given:

Mass of object, m = 100 lb

height fallen, h = 600 ft

initial velocity, u = 50 ft/s

acceleration due to gravity, g = 31.5 ft/s^2

Find final velocity when it touches ground.

Solution:

Use standard kinematics equation, in the absence of air resistance and variation of g with height,

v^2 - u^2 = 2aS

where

v = final velocity

u = initial velocity

a = acceleration due to gravity

S = distance travelled

Substitute values

v^2 = u^2 + 2aS

= 50^2 + 2*31.5*600

= 40300 ft^2/s^2

Final velocity,

v = sqrt(40300) ft/s

= 200.75 ft/s

= 201 ft/s  to the nearest foot.

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6