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SUBJECT: Basic Calculus
NOTE: Provide complete solution.

A stone is thrown upward from the top of a tower 50 meters high. Its velocity in an upward direction t second later is 20 - 5t meters per second. Find the height of the stone above the ground 3 seconds later.​

Answer :

semsee45

Answer:

87.5 meters

Step-by-step explanation:

To find the height of the stone above the ground 3 seconds after it has been thrown directly upward, we need to integrate the velocity function to obtain the displacement function (height function), and then evaluate it at t = 3.

Given that the velocity function is v(t) = 20 - 5t meters per second, the displacement function s(t) can be found by integrating the velocity function with respect to time (t):

[tex]\displaystyle s(t) = \int v(t) \; dt = \int (20 - 5t) \; dt[/tex]

Integrating term by term, we get:

[tex]s(t) = 20t - \dfrac{5}{2}t^2 + C[/tex]

where C is the constant of integration.

To determine the value of C, we can use the initial condition that the stone is thrown upward from the top of a tower 50 meters high. This means when t = 0, s = 50. Therefore:

[tex]s(0) = 50\\\\\\20(0) - \dfrac{5}{2}(0)^2 + C = 50\\\\\\C=50[/tex]

So, the displacement function is:

[tex]s(t) = 20t - \dfrac{5}{2}t^2 + 50[/tex]

Now, to find the height of the stone above the ground 3 seconds later, we evaluate s(t) at t = 3:

[tex]s(3) = 20(3) - \dfrac{5}{2}(3)^2 + 50\\\\\\s(3) = 60 - \dfrac{45}{2} + 50\\\\\\s(3) = 60 - 22.5 + 50\\\\\\s(3) = 87.5[/tex]

Therefore, the height of the stone above the ground 3 seconds after it has been thrown directly upward is 87.5 meters.

asuralite

Answer:

[tex]\Huge \boxed{\boxed{\boxed{\boxed{\bf{87.5\, m}}}}}[/tex]

Step-by-step explanation:

Introduction and given information

After the stone is hurled straight upward, we must integrate the velocity function to get the displacement function, or height function, and then evaluate it at t = 3 in order to determine the stone's height above the ground three seconds later.

Given:

  • The stone is thrown upward from the top of a tower 50 meters high.
  • The initial upward velocity of the stone is 20 m/s.
  • The velocity of the stone t seconds later is given by v = 20 - 5t m/s.

We need to find the height of the stone above the ground 3 seconds after it is thrown.

Solution

First, we can integrate the velocity function to obtain the displacement function (height function). The velocity function is v(t) = 20 - 5t m/s.

Integrating both sides:

  • [tex]\tt{s(t) = \int v(t) dt = \int (20 - 5t) dt}[/tex]
  • [tex]\boxed{\tt{s(t) = 20t - \frac{5}{2}t^2 + C}}[/tex]

Where:

  • s(t) is the displacement function.
  • C is the constant of integration.

Now, we should determine the value of the constant of integration C using the initial condition. At t = 0, the stone is at the top of the tower, so s(0) = 50 m.

Substituting t = 0 into the displacement function:

  • [tex]\tt{50 = 20(0) - \frac{5}{2}(0)^2 + C}[/tex]
  • [tex]\tt{C = 50}[/tex]

Next, write the complete displacement function

  • [tex]\boxed{\tt{s(t) = 20t - \frac{5}{2}t^2 + 50}}[/tex]

Now, evaluate the displacement function at t = 3 seconds to find the height above the ground

  • [tex]\tt{s(3) = 20(3) - \frac{5}{2}(3)^2 + 50}[/tex]
  • [tex]\tt{s(3) = 60 - \frac{45}{2} + 50}[/tex]
  • [tex]\tt{s(3) = 60 - 22.5 + 50}[/tex]
  • [tex]\tt{s(3) = 87.5}[/tex]

Therefore, the height of the stone above the ground 3 seconds after it has been thrown directly upward is 87.5 metres.

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