April shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high. The
pathway of the arrow can be represented by the equation h = -16t
2 + 80t + 25, where h is the
height and t is the time in seconds. What is the maximum height of the arrow?
Answers
Given : shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high.
The pathway of the arrow can be represented by the equation
h = -16t² + 80t + 25, where h is the height and t is the time in seconds.
To Find : the maximum height of the arrow
Solution:
h = -16t² + 80t + 25
dh/dt = -32t + 80
dh/dt = 0
=> -32t + 80 = 0
=> 2t - 5 = 0
=> t = 5/2
d²h/dt² = -32 < 0
hence maximum value
maximum height is at t = 5/2
Substitute t = 5/2 in
h = -16t² + 80t + 25
=> h = -16(5/2)² + 80(5/2) + 25
= -100 + 200 + 25
= 125
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Step-by-step explanation:
Given : shoots an arrow upward at a speed of 80 feet per second from a platform 25 feet high.
The pathway of the arrow can be represented by the equation
h = -16t² + 80t + 25, where h is the height and t is the time in seconds.
To Find : the maximum height of the arrow
Solution:
h = -16t² + 80t + 25
dh/dt = -32t + 80
dh/dt = 0
=> -32t + 80 = 0
=> 2t - 5 = 0
=> t = 5/2
d²h/dt² = -32 < 0
hence maximum value
maximum height is at t = 5/2
Substitute t = 5/2 in
h = -16t² + 80t + 25
=> h = -16(5/2)² + 80(5/2) + 25
= -100 + 200 + 25
= 125
Learn More:
A particle is projected in the upword direction under gravity time of ...
brainly.in/question/11383332
11. A particle is projected vertically upward and movesfreely under ...
brainly.in/question/12773764
