Physics, 25.07.2019 14:00 QueenBlumple4443

Astone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9.8 m/s2.) (a) find the distance (in meters) of the stone above ground level at time t. h(t) = −4.9t2+750 (b) how long does it take the stone to reach the ground? (round your answer to two decimal places.) 12.37 s (c) with what velocity does it strike the ground? (round your answer to one decimal place.) -121.2 m/s (d) if the stone is thrown downward with a speed of 6 m/s, how long does it take to reach the ground? (round your answer to two decimal places.)

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Answer from: zitterkoph

(a) The stone moves by uniform accelerated motion, with constant acceleration g=9.81 m/s^2

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
$y(t)= y_0 - \frac{1}{2}gt^2= 750 - 4.9 t^2$

(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
750-4.9 t^2 = 0

from which we find the time t after which the stone reaches the ground:
$t= \sqrt{\frac{750 m}{4.9 m/s^2 }}= 12.37 s$

(c) The velocity of the stone at time t can be written as
v(t) = -gt

because it is an accelerated motion with initial speed zero. Substituting t=12.37 s, we find the final velocity of the stone:
v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s

v(12.37 s)=-(9.81 m/s^2)(12.37 s)=-121.3 m/s

(d) if the stone has an initial velocity of v_0 = 6 m/s

, then its law of motion would be
$y(t)=y_0 - v_0t - \frac{1}{2}gt^2$
and we can find the time it needs to reach the ground by requiring again y(t)=0:
0=750 - 6t - 4.9 t^2

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.

Answer from: rileybaby34

a) $\rm y(t) = 750-4.9t^2$

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,

$\rm y(t) = y_0 - \dfrac{1}{2}gt^2$

At t = 0 , y = 750 m

$\rm y(t) = 750 - \dfrac{1}{2}\times 9.81\times t^2$

$\rm y(t) = 750-4.9t^2$ --- (1)

b) The stone is reached to the ground that is, y(t) = 0

So, from equation (1) we get

$\rm t = \sqrt{\dfrac{750}{4.9}}$

t = 12.37 sec

c) Velocity when it strike the ground is given by,

v = u - gt

here initial velocity u = 0, therefore

$\rm v = - 9.81 \times 12.37$

v = - 121.34 m/sec

d) In this case given that initial velocity is 6 m/sec, therefore

$\rm y(t) = y_0 - ut-\dfrac{1}{2}gt^2$ ---- (2)

here stone is reached to the ground so y(t) = 0. From equation (2) we get,

$\rm 0 = 750-(6\times t) -\dfrac{1}{2}\times 9.81 \times t^2$

$\rm 4.9t^2 + 6t -750 = 0$

By solving the above equation we get t = 11.78 sec.

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Answer from: Quest

So what is the question

Answer from: Quest

yo yo yo its papaguy in the house lol

okay, ima .

um it shows links to uh so kinda microsoft website is that were ima go or no? ?

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Astone is dropped from the upper observation deck of a tower, 750 m above the ground. (assume g = 9....