An object is thrown vertically from a height of #12 m# at # 1 m/s#. How long will it take for the object to hit the ground?

Answer 1

time #t=1.67028567" "#seconds

#y=v_yt+1/2*g*t^2#
#-12=1*t+1/2*-9.8*t^2#
#-24=2t-9.8t^2#
#9.8t^2-2t-24=0#
#4.9t^2-t-12=0#
#t=(-(-1)+-sqrt((-1)^2-4(4.9)(-12)))/(2*(4.9))#
#t=(+1+-sqrt(1-4(4.9)(-12)))/(2*(4.9))#
#t=1.67028567" "#seconds

Blessings...I hope this clarification is helpful.

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Answer 2

#2# seconds

One of the kinematic equations can be used to find the time it takes for the object to hit the ground, assuming that it is thrown upwards and then falls.

We will use positive values to indicate a downward direction for the given problem.

#Deltad=12m#
#v_i=-1m/s#
#a=9.81m/s^2#
#Deltat=?#
Use the kinematic equation, #Deltad=v_iDeltat+1/2aDeltat^2#, to solve for #Deltat#.
#Deltad=v_iDeltat+1/2aDeltat^2#
#0=1/2aDeltat^2+v_iDeltat-Deltad#

Enter the values.

#0=color(darkorange)(1/2(9.81m/s^2))Deltat^2+(color(teal)(-1m/s))Deltat# #color(violet)(-12m)#
Use the quadratic formula to solve for #Deltat#.
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-(color(teal)(-1m/s))+-sqrt((color(teal)(-1m/s))^2-4(color(darkorange)(1/2(9.81m/s^2))(color(violet)(-12m)))))/(2(color(darkorange)(1/2(9.81m/s^2))))#
#x=1.67scolor(white)(i),color(white)(i)color(red)cancelcolor(black)(-1.47s)#
Since we are looking for the time it takes for the object to hit the ground after it is thrown, we take the positive value, #1.67s#, ignore #-1.47s#.
By rounding #1.67s# to #1# significant digit, it becomes #2s#, which is the time it takes for the object to hit the ground.
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Answer 3

To solve for the time it takes for the object to hit the ground, you can use the equation:

[ t = \sqrt{\frac{2h}{g}} ]

where:

  • ( t ) is the time taken,
  • ( h ) is the initial height (12 m in this case),
  • ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )).

Plugging in the values:

[ t = \sqrt{\frac{2 \times 12 , \text{m}}{9.8 , \text{m/s}^2}} ]

[ t ≈ \sqrt{\frac{24}{9.8}} ]

[ t ≈ \sqrt{2.45} ]

[ t ≈ 1.57 , \text{s} ]

So, it will take approximately 1.57 seconds for the object to hit the ground.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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