1. May 25, 2005

### denafoster

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?

2. May 25, 2005

### arildno

What is the height value corresponding to ground level?

3. May 25, 2005

### denafoster

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. May 25, 2005

### arildno

Both of these earlier answers are correct.

5. May 25, 2005

### denafoster

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. May 25, 2005

### arildno

7. May 25, 2005

### denafoster

It would be zero.

8. May 25, 2005

### arildno

Correct.
So what equation must the instant "t" fulfill which coresponds to the dart hitting the ground?

9. May 25, 2005

### denafoster

I'm guessing that something needs to be set to equal zero . . . but forgive me, I'm still a little lost.

10. May 25, 2005

### arildno

You know that height as a function of time is given by the equation:
$$d(t)=-4.9t^{2}+20t+1.5 (1)$$
Now, that d=0 (that is, we have reached groundlevel), means that the instant "t" corresponding to that must fulfill:
$$0=-4.9t^{2}+20t+1.5 (2)$$

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):
$$d(T)=-4.9T^{2}+20T+1.5 (3)$$
c) a) says now that d(T)=0, and inserting this insight into (3) yields:
$$0=d(T)=-4.9T^{2}+20T+1.5$$
That is:
$$0=-4.9T^{2}+20T+1.5(4)$$

(4) can now be solved for T, remembering that T must be greater than zero.

Last edited: May 25, 2005
11. May 25, 2005