Suppose Neil Armstrong decided to throw a golf ball into the air while he was standing on the moon and that the height of the golf ball was modeled by the equation below, where s is
measured in feet and t is measured in seconds s(t) = -2.72t2 + 26.9t + 6. Find the best
approximation for the instantaneous rate of change (velocity) of the golf ball at 7 seconds.

At 7 seconds the instantaneous rate of change of the function s(t) with time which is the velocity is -11.18 m/s

Step-by-step explanation:

The given information on the height can be written as follows;

s(t) = -2.72·t² + 26.9·t + 6

Where;

s = The height (displacement) of the golf ball

t = The time taken during the motion of the golf ball

The velocity is found by finding the derivative of the displacement s as follows;

v = d(s(t))/dt = d(-2.72·t² + 26.9·t + 6)/dt = 2×(-2.72)×t + 26.9

When t = 7, v = 2×(-2.72)×7 + 26.9 = -11.18 m/s

Therefore the ball is coming down with 11.18 m/s speed at 7 seconds.

sina-sinb

cosb+cosa

tana-tanb

cos0+cos x pi/3

20tan

step-by-step explanation: