Question

The height (feet) of a ball launched in the air t seconds after it is launched is given by: f(t) = - 16t^2 + 299t + 9. Find the height of the ball when it is falling at 110 ft/sec. The height is feet. If necessary, round to two decimal places. If there is more than one correct answer, separate them with commas.

          The height (feet) of a ball launched in the air t seconds after it is launched is given by:
f(t) = - 16t^2 + 299t + 9.
Find the height of the ball when it is falling at 110 ft/sec.
The height is feet.
If necessary, round to two decimal places. If there is more than one correct answer, separate them with commas.

The height (feet) of a ball launched in the air t seconds after it is launched is given by:
f(t) = - 16t^2 + 299t + 9.
Find the height of the ball when it is falling at 110 ft/sec.
The height is feet.
If necessary, round to two decimal places. If there is more than one correct answer, separate them with commas.

Added by Domingo I.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Satyam Gupta

10/14/2021

Step 1

The velocity of the ball is the derivative of the height function f(t) with respect to time t. Show more…

Show all steps

Thanks for your feedback!

The height (feet) of a ball launched in the air t seconds after it is launched is given by: f(t)=-16t^(2)+299t+9 Find the height of the ball when it is falling at 110f(t)/(sec). The height is feet. If necessary, round to two decimal places. If there is more than one correct answer, separate them with commas. The height (feet of a ball launched in the air t seconds after it is launched is given by f(t)= - 16t2+299t+9. Find the height of the ball when it is falling at 110 ft/sec. The height is feet.