A ball is shot from a cannon into the air with an upward velocity of 65 ft/sec. The equation that gives the height (h) of the ball at any time (t) is: h(t)= -14t^2+65t+1.75.

Find the maximum height attained by the ball.

What time was the maximum height achieved?

How high was the ball off of the ground when it was shot?

How far did the ball travel before it hit the ground

Step-by-step explanation:

Find AOS

X = -b/2a

X = -65/2(-14)

X = -65/-28

X = 2.32 secs

t = 2.32secs

A) y value when t = 2.32 is the maximum height.

Substitute t in the equation.

h(t)= -14t^2+65t+1.75

h(t)= (-14 * 2.32²) + (65 * 2.32) + 1.75

= (-14 * 5.38 ) + (65 * 2.32) + 1.75

= -75.35 + 150.8 + 1.75

= 77.2 ft

Maximum height attained by the ball = 77.2 ft

B) Time the maximum height achieved = 2.32secs

C) The ball off of the ground when it was shot is well t = 0

Substitute t in equation

h(t)= -14t^2+65t+1.75

h(0)= (-14 * 0²) + (65 * 0) + 1.75

= 0 + 0 + 1.75ft

The ball is 1.75ft off the ground when it was shot.

on your paper it doesn't say the answer but the the answer is altitude

step-by-step explanation:

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)