Respuesta :
(a) The stone moves by uniform accelerated motion, with constant acceleration [tex]g=9.81 m/s^2[/tex] directed downwards, and its initial vertical position at time t=0 is 750 m. So, the vertical position (in meters) at any time t can be written as
[tex]y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2[/tex]
(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
[tex]750-4.9 t^2 = 0[/tex]
from which we find the time t after which the stone reaches the ground:
[tex]t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s[/tex]
(c) The velocity of the stone at time t can be written as
[tex]v(t) = -gt[/tex]
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
[tex]v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s[/tex]
(d) if the stone has an initial velocity of [tex]v_0 = 6 m/s[/tex], then its law of motion would be
[tex]y(t)=y_0 - v_0t - \frac{1}{2}gt^2 [/tex]
and we can find the time it needs to reach the ground by requiring again y(t)=0:
[tex]0=750 - 6t - 4.9 t^2 [/tex]
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.
[tex]y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2[/tex]
(b) The time the stone takes to reach the ground is the time at which the vertical position of the stone becomes zero: y(t)=0. So, we can write
[tex]750-4.9 t^2 = 0[/tex]
from which we find the time t after which the stone reaches the ground:
[tex]t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s[/tex]
(c) The velocity of the stone at time t can be written as
[tex]v(t) = -gt[/tex]
because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
[tex]v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s[/tex]
(d) if the stone has an initial velocity of [tex]v_0 = 6 m/s[/tex], then its law of motion would be
[tex]y(t)=y_0 - v_0t - \frac{1}{2}gt^2 [/tex]
and we can find the time it needs to reach the ground by requiring again y(t)=0:
[tex]0=750 - 6t - 4.9 t^2 [/tex]
which has two solutions: one is negative so we neglect it, while the second one is t=11.78 s, so this is the time after which the stone reaches the ground.
a) [tex]\rm y(t) = 750-4.9t^2[/tex]
b) t = 12.37 sec
c) v = -121.34 m/sec
d) t = 11.78 sec
Given :
Upper observation deck of a tower is 750 m above the ground.
Solution :
a) The vertical position at any time t is given by,
[tex]\rm y(t) = y_0 - \dfrac{1}{2}gt^2[/tex]
At t = 0 , y = 750 m
[tex]\rm y(t) = 750 - \dfrac{1}{2}\times 9.81\times t^2[/tex]
[tex]\rm y(t) = 750-4.9t^2[/tex] --- (1)
b) The stone is reached to the ground that is, y(t) = 0
So, from equation (1) we get
[tex]\rm t = \sqrt{\dfrac{750}{4.9}}[/tex]
t = 12.37 sec
c) Velocity when it strike the ground is given by,
v = u - gt
here initial velocity u = 0, therefore
[tex]\rm v = - 9.81 \times 12.37[/tex]
v = - 121.34 m/sec
d) In this case given that initial velocity is 6 m/sec, therefore
[tex]\rm y(t) = y_0 - ut-\dfrac{1}{2}gt^2[/tex] ---- (2)
here stone is reached to the ground so y(t) = 0. From equation (2) we get,
[tex]\rm 0 = 750-(6\times t) -\dfrac{1}{2}\times 9.81 \times t^2[/tex]
[tex]\rm 4.9t^2 + 6t -750 = 0[/tex]
By solving the above equation we get t = 11.78 sec.
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