If an object is dropped from a tower, then the velocity, V (in feet per second), of the object after t seconds can be obtained by multiplying t by 32 and adding 10 to the result. Find V as a linear function of t, and use this function to evaluate V (7.8), the velocity of the object at time t = 7.8 seconds.

A. V (7.8) = 257.6 ft per sec

B. V (7.8) = 260.9 ft per sec

C. V (7.8) = 259.6 ft per sec

D. V (7.8) = 258.9 ft per sec

E. None of these

259.6 ft/sec

Step-by-step explanation:

it is a universal standard that acceleration due to gravity is 32ft/sec^2.

Now it can be verified by equation,

V (f) = V (i) + at (1st equation of motion derived by Newton's three laws of motion)

where,

V (f) is final velocity

V (i) is initial velocity

a is acceleration which is constant and have value 32ft/sec^2

t is time which is given as 7.8 seconds

In the given case, initial velocity that is V (i) will be 0ft/sec. Because, on dropping, object will start to move under the influence of gravity from zero speed.

So,

V (f) = 0 + (32) (7.8)

V (f) = 249.6 ft/sec

Now the condition is given that you have to add a constant 10 to the answer.

so, V (f) = 249.6 + 10

V (f) = 259.6 ft/sec