Question

A ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 10 ft/sec. Its position function is s(t)=−16t2+10t+100.

a. What is its velocity in ft/sec when t = 2 seconds? (Solve by using instantaneous rate.)
b. Determine the equation of a line, in slope-intercept form, that passes through the points (5, 6) and (10, 2).

Answer 1

Answer;

-54ft/s

y = -4/5 x + 10

Explanation

Given the position of an object expressed by the function

s(t)=−16t²+10t+100

Velocity is the change in position with respect to time

v(t) = ds(t)/dt

v(t) = -32t + 10

When t = 2

v(2) = -32(2)+10

v(2) = -64+10

v(2) = -54

Hence the velocity of the object is -54ft/s

b) The standard equation of a line in point slope form is expressed as;

y = MX+c

M is the slope

c is the y-intercept

Given the coordinate (5, 6) and (10, 2)

M = 2-6/10-5

M = -4/5

Get the y-intercept

Substitute m = -4/5 and any point say (5,6) into the expression y = mx+c

6 = -4/5 (5) + c

6 = -4+c

c = 6+4

c = 10

Get the required equation

Recall that: y = mx+c

y = -4/5 x + 10

Hence the equation of a line, in slope-intercept form is y = -4/5 x + 10