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A projectile with an initial velocity of 48 feet per second is launched from a building 190 feet tall. The path of the projectile is modeled using the equation h(t) = â€""16t2 48t 190. Approximately when will the projectile hit the ground? 1. 5 seconds 3. 2 seconds 5. 3 seconds 6. 2 seconds.

Respuesta :

onyebuchinnaji

The time for the projectile to hit the ground from the given position is 1.5 s.

The given parameters:

  • Initial velocity of the projectile, u = 48 ft/s
  • Height, h = 190 ft
  • Model of the projectile path, h(t)  = –16t² + 48t + 190.

The time for the projectile to hit the ground is calculated as follows:

when the projectile hits the ground, the final velocity = 0

[tex]v_f = 0\\\\\frac{dh(t)}{dt} = v_f\\\\\frac{dh(t)}{dt} = -32t + 48\\\\-32t+48 = 0\\\\32t = 48\\\\t = \frac{48}{32} \\\\t = 1.5 \ s[/tex]

Thus, the time for the projectile to hit the ground from the given position is 1.5 s.

Learn more about time of motion of projectiles here: https://brainly.com/question/13533552

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