If a ball is thrown upward at 28.6 meters per second from the top of a building that is 51 meters high, the height of the ball can be modeld by s (t) = 51 + 28.6t – 4.9t^2 meters, where tis the number of seconds after the ball is thrown. What is the ball's maximum height?

Orion Cervantes

Orion Cervantes

Answered question

2022-09-21

If a ball is thrown upward at 28.6 meters per second from the top of a building that is 51 meters high, the height of the ball can be modeld by s ( t ) = 51 + 28.6 t 4.9 t 2 meters, where tis the number of seconds after the ball is thrown. What is the ball's maximum height?

Answer & Explanation

Matthias Calhoun

Matthias Calhoun

Beginner2022-09-22Added 11 answers

When ball is at maximum height
d 5 ( t ) d t = 0
given that
s ( t ) = 51 + 28 6 t 4 9 t 2 d s ( t ) d t = 0 28 6 4 9 × 2 t = 0 9 8 t = 28 6 t = 28 6 9 8
When t = 886 98 sec. ball will be 91 maximum height
= 51 + 28 6 × 286 98 4 9 × ( 286 98 ) 2 = 92.7326531
maximum height
=92.7 m

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