April shoots an arrow upward at a speed of 80 ft/sec from a platform of 25 ft. 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 that the arrow reaches

125feet

Step-by-step explanation:

Given the equation that modeled the height expressed as h = -16t^2 + 80t + 25, where h is the height and t is the time in seconds.

The arrow reaches the maximum height at dh/dt = 0

dh/dt = -32t + 80

0= -32t+80

32t = 80

t = 80/32

t = 2.5secs

substitute t = 2.5 into the formula;

h = -16t^2 + 80t + 25

h = -16(2.5)^2 + 80(2.5) + 25

h = -16(6.25)+225

h = -100+225

h = 125

Hence the maximum height the arrow reaches is 125feet

sup

step-by-step explanation:

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: