Question

10. A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is $s(t) =$ $96 + 80t - 16t^2$. a. After how many seconds does the ball reaches the maximum height. b. What is the maximum height?

          10. A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of
80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is $s(t) =$
$96 + 80t - 16t^2$.
a. After how many seconds does the ball reaches the maximum height.
b. What is the maximum height?

10. A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of
80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t) =
96 + 80t - 16t^2.
a. After how many seconds does the ball reaches the maximum height.
b. What is the maximum height?

Added by Jonathan M.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Penny Riley

01/03/2023

Step 1

Step 1: To find the time it takes for the ball to reach the maximum height, we can use the formula for the time it takes for an object to reach its maximum height when thrown vertically upward: t = -b/2a, where a = -16 (from the coefficient of t^2) and b = 80 (from Show more…

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10. A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of 80 feet per second. The distance s (in feet) of the ball from the ground after t seconds is s(t) = 96 +80t - 16t². a. After how w many seconds does the ball reaches the maximum height. b. What is the maximum height? 10. A ball is thrown vertically upward from the top of a building 96 feet tall with an initial velocity of 80 feet per second.The distance s in fect of the ball from the ground after t seconds is st= 96+80t-16t2. a.After how many seconds does the ball reaches the maximum height b.What is the maximum height?