 , 22.06.2019 21:00 reyrey216

# If an object is projected upward with an initial velocity of 126ft per? sec, its height h after t seconds is h=? 16t^2+126t. find the height of the object after 4 seconds =760 ft

Step-by-step explanation:

h=16t^2+126t

Let t=4

h = 16 (4)^2 +126(4)

h = 16*16 +126(4)

=256+504

=760 ft 236ft.

Step-by-step explanation: To solve the given problem, we need to substitute the given time (4 seconds) in the equation to calculate the height of the object with an initial velocity of 129ft per sec. The substitution is shown as follow:  h=-256+492

h=236ft.

The height of the object after 4 seconds is 236ft. 8 centermeters

Step-by-step explanation: This problem can be solved directly by neglecting the initial velocity given. All we have to do is to plug in t = 4 in the given equation. That is:

h = -16^2 + 123 * 4

h = 236 ft You are given a function This function shows you the dependence of height from time.

To find the height of the object after 4 seconds, you should substitute t=4 into the function expression: the height of the object after 4 seconds will be 252 ft. After 4 seconds
h(4) = -16(4)^2 + 126(4) = -16(16) + 504 = -256 + 504 = 248 feet. I would think it is
h=-16t^2+124t
after 4 seconds or when t=4 find h
h=-16(4^2)+124(4)
h=-16(16)+496
h=-256+496
h=240

height is 240 feet This equation should be written h(t) = -16t^2 + 129t.  After 4 sec, the height of this object is

h(4) = -16(4)^2 + 129(4) = -256 + 516 = 260 feet. To find the height of the object after 4 seconds, just input the 4 where t is located.

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If an object is projected upward with an initial velocity of 126ft per? sec, its height h after t s...

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