serenah99
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Crickets can jump with a vertical velocity of up to 14 ft/s. Which equation models the height of such a jump, in feet, after t seconds? h(t) = –16t2 + v0t + h0

h(t) = –16t2 + v0t + 14

h(t) = –16t2 + 14t

h(t) = –16t2 + 14t + 14

Respuesta :

irspow
acceleration is simplified by assuming it is the constant -g

a=-g  we integrate this with respect to time to get v...

v=-gt+C  where C is the initial velocity in this case 14ft/s so

v=-gt+14  integrate again to get the height function

h=(-gt^2)/2 +14t +C  we are not given an initial height so C is 0

h(t)=14t-gt^2/2   letting g=32 and neatening up a bit...

h(t)=14t-16t^2

Answer:

Option 3. h(t) = -16t² + 14t

Step-by-step explanation:

As we know the formula of motion under gravity is

h = ut - (1/2)gt²

Here h is the height covered by the cricket with vertical velocity u and (-g) is the gravitational pull downwards.

Since the value of g = 9.8 mt/sec²

It can be written in feet as g = 32 feet/sec²

Vertical velocity of the crickets u = 14 ft/sec

Now by putting these values in the formula

h = 14t - (1/2)×32×t²

h = 14t - 16t²

we can rewrite this equation in the form of function

h(t) = -16t² + 14t

So the answer is Option 3. h(t) = -16t² + 14t

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