A stone is launched into the air from a height of 240 feet. The height, h, of the stone, in feet, t seconds after launch is given by
the formula h = 16t^2 + 32t+ + 240. After how long will the stone hit the ground?

A stone is launched into the air from a height of 240 feet The height h of the stone in feet t seconds after launch is given by the formula h 16t2 32t 240 After class=

Respuesta :

A projectile is an object launched into space, and moving under the influence of gravity and momentum. The stone here is an example. Thus, the time taken for the stone to reach the ground is 10 seconds.

On launching of the stone into space, it behaves as a projectile. Thus moving under the influence of gravity and its momentum. Its momentum is the product of its mass and velocity.

The height h, is given as:

h = 16[tex]t^{2}[/tex] + 32t + 240

Differentiating the expression with respect to t, we have;

32t + 32 = 0

t = -1 seconds

Substitute the value of time, t, in the equation for the height;

h = 16 - 32 + 240

  = 224

h = 224 m

Total height, S = 240 + 224

                         = 464

S = 464 m

The total height covered by the stone is 464 m.

From the second equation of free fall,

S = Ut + [tex]\frac{1}{2}[/tex]g[tex]t^{2}[/tex]

S = [tex]\frac{1}{2}[/tex]g[tex]t^{2}[/tex]

464 = [tex]\frac{1}{2}[/tex]*9.8*[tex]t^{2}[/tex]

[tex]t^{2}[/tex] = [tex]\frac{928}{9.8}[/tex]

  = 94.694

t = [tex]\sqrt{94.694}[/tex]

 = 9.7311

t 10 seconds

The time taken for the stone to hit the ground is 10 seconds.

Thus option A is the required answer.

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