Respuesta :
Answer
given,
height of the building = 160 ft
initial velocity = 48 ft/s
the relation of distance is given
d = -16 t²  + 48 t + 160
time at which the ball hit the ground will be when d = 0
-16 t²  + 48 t + 160 = 0
solving quadratic equation
[tex]t = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]t = \dfrac{-48\pm \sqrt{48^2+4\times 16 \times 160}}{2\times (-16)}[/tex]
[tex]t = \dfrac{-48\pm 112}{-32}[/tex]
[tex]t = \dfrac{-48+ 112}{-32},\dfrac{-48- 112}{-32}[/tex]
t = 5 s(neglecting the negative solution)
b) ball will reach max height when velocity is equal to zero
 [tex]v = \dfrac{dd}{dt}[/tex]
 [tex]v = \dfrac{d}{dt}(-16 t^2+48 t +160)[/tex]
 [tex]v = -32 t + 48[/tex]
  v = 0
   t = 1.5 s
at 1.5 sec the ball will reach at maximum height
A) The number of seconds that it will take for the ball to strike the ground is; t = 5 seconds.
B) The time it will take for the ball to reach its' maximum height is;
t = 1.5 seconds
We are given the distance equation of the ball from the ground after t seconds as; d(t) = 160 + 48t – 16t²
A) The ball will strike the ground at d(t) = 0. Thus;
160 + 48t – 16t² = 0
From online quadratic equation solver, we have the solution to the equation as; t = -2 or t = 5
time cannot be negative and so we will adopt t = 5 s.
Thus, The number of seconds that it will take for the ball to strike the ground is 5 seconds.
B) The maximum height will be reached when the velocity is zero i.e. when the derivative of the distance equation is zero.
Thus;
d'(t) = 48 - 32t
At d'(t) = 0
48 - 32t = 0
32t = 48
t = 48/32
t = 1.5 seconds
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