An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it falls back to the ground. a. What height is the object thrown from? b. How long does it take for the object to hit the ground?

Answers

Answer 1

Answer:

The value is [tex]T_t = 2.5659 \ s[/tex]    

Explanation:

From the we are told that  

         The initial speed of the object is  [tex]u = 8 \ m/s[/tex]

         The greatest height it reached is  [tex]h = 15 \ m[/tex]

Generally from kinematic equation we have that

      [tex]v^2 = u^2 + 2gH[/tex]

At maximum height v  =  0 m/s

So

      [tex]0^2 = 8^2 + 2 * - 9.8 * H[/tex]

=>    [tex]H = 3.27 \ m[/tex]

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      [tex]s = h - H[/tex]

=>    [tex]s = 15 - 3.27[/tex]

=>    [tex]s = 11.73 \ m[/tex]

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            [tex]v = u + (-g) t[/tex]

At maximum height v  =  0 m/s

           [tex]0 = 8 - 9.8t[/tex]

=>         [tex]t = 0.8163 \ s[/tex]

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       [tex]h = ut_1 + \frac{1}{2} gt_1^2[/tex]

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       [tex]15 = 0* t + 4.9 t^2[/tex]

=>      [tex]t_1 = 1.7496[/tex]

Generally the total time taken is mathematically represented as

          [tex]T_t = t + t_1[/tex]

=>        [tex]T_t = 0.8163 + 1.7496[/tex]

=>        [tex]T_t = 2.5659 \ s[/tex]            

 


Related Questions

Electric power is to be generated by installing a hydraulic turbine-generator at a site 70 m below the free surface of a large water reservoir that can supply water at a rate of 1500 kg/s steadily. If the mechanical power output of the turbine is 800 kW and the electric power generation is 750 kW, determine the turbine efficiency and the combined turbine–generator efficiency of this plant. Neglect losses in the pipes.

Answers

Answer:

[tex]\eta_{turbine} = 0.777 = 77.7\%[/tex]

[tex]\eta_{combined} = 0.728 = 72.8\%[/tex]

Explanation:

First we calculate the power input to the turbine. The input power will be equal to the potential energy of water per unit time:

Input Power = [tex]P_{in} = \frac{Work}{Time} = \frac{Potential\ Energy\ of\ Water}{t} \\P_{in} = \frac{(mass)(g)(height)}{Time} = (mass flow rate)(g)(height)\\\\P_{in} = (1500\ kg/s)(9.81\ m/s^2)(70\ m)\\P_{in} = 1.03\ x\ 10^6\ W = 1030\ KW[/tex]

Now, for turbine efficiency:

[tex]\eta_{turbine} = \frac{Mechanical\ Power\ Out}{P_{in}}\\\\\eta_{turbine} = \frac{800\ KW}{1030\ KW}\\\\\eta_{turbine} = 0.777 = 77.7\%[/tex]

for generator efficiency:

[tex]\eta_{generator} = \frac{Power\ Generation}{Mechanical\ Power\ Out}\\\\ \eta_{generator} = \frac{750\ KW}{800\ KW}\\\\\eta_{turbine} = 0.9375 = 93.75\%[/tex]

Now, for combined efficiency:

[tex]\eta_{combined} = \eta_{turbine}\ \eta_{generator}\\\\\eta_{combined} = (0.777)(0.937)\\\eta_{combined} = 0.728 = 72.8\%[/tex]

How do I calculate how many meters are in 7.2 light years?
The exact question is:
Calculate in meters the distance between a galaxy and the Earth if the distance is equal to 7.2 light years.

Answers

Answer:

68.2 Quadrillion meters

Explanation:

A lightyear is the distance that light travels in one year.

Speed of light is [tex]3*10^8\ m/s[/tex]

So light covers 300,000,000 meters in one second.

One year has 31536000 seconds so , light covers

[tex]9.461*10^{15}\ meters\ in\ one\ year[/tex]

so 7.2 light years is

[tex]7.2*(9.461*10^{15})\\6.82*10^{16}[/tex]

so 7.2 light years is

6.82 x 10^(16) meters or

68.2 Quadrillion meters

Help please!!!
if superman at 90kg jumps a 40m building in a single bound how much work does superman perform

Answers

Answer:

5 years worth of work (aka all of the homework i currently have)

An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?

Answers

Answer:

Explanation:

Given parameters:

Weight of object  = 49N

Force applied = 12N

Unknown:

Acceleration of object  = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

 Acceleration  = [tex]\frac{12}{49}[/tex]   = 0.25m/s²

what power is transmitted by 2A flowing across 5V​

Answers

Answer:

[tex]\huge\boxed{10\:watts}[/tex]

Explanation:

Power = Current × Voltage

P = I × V

P = 2 × 5

P = 10 watts

Help me pretty please with a cherry on top!!

Answers

it doesnt show anything for me..


explaination :: cornje

Answer:

A. im not for sure hope it helps

Explanation:

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