SOLUTION: Solve using derivatives - A dart is shot straight up from 1.5 metres above ground level. The distance (d), in metres, the dart is above the ground at time (t), in seconds is:

Question 861872: Solve using derivatives - A dart is shot straight up from 1.5 metres above ground level. The distance (d), in metres, the dart is above the ground at time (t), in seconds is:

1) What is the maximum height the dart reaches? I think this is 21.92 m
2) for how many seconds is the dart in the air?
Found 3 solutions by ewatrrr, cigany29, Alan3354:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi

took the derivative as directed and set to zero to find x-value at that point
-9.8t + 20 = 0, t = -20/9.8 = 2.0408
substituted that x-value found into d(t) to find the height

Answer by cigany29(23)   (Show Source): You can put this solution on YOUR website!

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Solve using derivatives - A dart is shot straight up from 1.5 metres above ground level. The distance (d), in metres, the dart is above the ground at time (t), in seconds is:

1) What is the maximum height the dart reaches? I think this is 21.92 m
Already answered.
-------------
2) for how many seconds is the dart in the air?
From t = 0 until d(t) = 0 at impact.

Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=429.4 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: -0.0736703080011585, 4.15530296106238. Here's your graph:

=============
Ignore the negative solution.
t =~ 4.155 seconds
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