Find the Flight Time of a Dart Shot from 1.5m Above Ground | Derivative Question

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In summary, the dart is shot straight up from 1.5m above ground level and reaches a height of 21.9m above ground level at the instant it hits the ground.
  • #1
denafoster
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A dart is shot straight up from 1.5m above ground level. The distance d,
in meters, the dart is above ground at time t, in seconds, is :

d(t) = -4.9t^2 + 20t + 1.5

For how many seconds is the dart in the air?
 
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  • #2
What is the height value corresponding to ground level?
 
  • #3
There was a previous part to the question which I have the answer for :

What is the velocity of the dart one second after it is shot upward?

v'(t) = -9.8t + 20 = -9.8(1) + 20 = 10.2 m/s

What is the Maximum height that the dart reaches ?

velocity will be zero at 2.0408 seconds [0=(-9.8)t + 20]

d(2.0408) = -4.9(2.0408)^2 + 20(2.0408) + 1.5 = 21.9 m
 
  • #4
Both of these earlier answers are correct.
 
  • #5
I am stuck on the final part of the question, I know how to solve it using a physics formula . . . but not with the derivative.
 
  • #6
Did you read post 2?
 
  • #7
It would be zero.
 
  • #8
Correct.
So what equation must the instant "t" fulfill which coresponds to the dart hitting the ground?
 
  • #9
I'm guessing that something needs to be set to equal zero . . . but forgive me, I'm still a little lost.
 
  • #10
You know that height as a function of time is given by the equation:
[tex]d(t)=-4.9t^{2}+20t+1.5 (1)[/tex]
Now, that d=0 (that is, we have reached groundlevel), means that the instant "t" corresponding to that must fulfill:
[tex]0=-4.9t^{2}+20t+1.5 (2)[/tex]

Note that we have been sloppy with our notation here:
In (1), "t" is used as a VARIABLE, whereas in (2), "t" is used as a fixed VALUE we're supposed to find.
If you want to be careful in your notation, proceed as follows:
a) Let T be the instant when the dart hits the ground. At that instant, the height value "d" is 0.
b) The value of the height at time T is given by evaluating our height formula at T (T is therefore an element in the interval over which the variable "t" ranges):
[tex]d(T)=-4.9T^{2}+20T+1.5 (3)[/tex]
c) a) says now that d(T)=0, and inserting this insight into (3) yields:
[tex]0=d(T)=-4.9T^{2}+20T+1.5[/tex]
That is:
[tex]0=-4.9T^{2}+20T+1.5(4)[/tex]

(4) can now be solved for T, remembering that T must be greater than zero.
 
Last edited:
  • #11
Thanks for your help today. :smile:
 
  • #12
Welcome to PF!
 

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function with respect to one of its variables. It is commonly used in calculus to find the slope of a tangent line at a specific point on a curve.

2. Why are derivatives important?

Derivatives are important because they allow us to analyze the behavior of complex functions and make predictions about their future values. They also have many practical applications in fields such as physics, engineering, finance, and economics.

3. How do you find the derivative of a function?

The derivative of a function can be found by using the rules of differentiation, which involve algebraic manipulation and taking limits. These rules allow us to find the derivative of any polynomial, exponential, logarithmic, or trigonometric function.

4. What is the difference between a derivative and an antiderivative?

A derivative represents the instantaneous rate of change of a function, while an antiderivative represents the original function before it was differentiated. In other words, derivatives are used to find the slope of a function, while antiderivatives are used to find the original function itself.

5. Can derivatives be negative?

Yes, derivatives can be negative. The sign of a derivative depends on the direction of the slope of a function at a specific point. If the slope is positive, the derivative will be positive, and if the slope is negative, the derivative will be negative.

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