# 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.

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.

h(4) = -16(4)^2 + 126(4) = -16(16) + 504 = -256 + 504 = 248 feet.

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

h(4) = -16(4)^2 + 129(4) = -256 + 516 = 260 feet.

So

Now just solve for h