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We can see from the graph that the toss is not strong enough to reach the level of the window.
To confirm this, let's find the vertex of the parabolic path.
We know that this vertex is on the axis of symmetry.
Substitute values
a(- b)=- a * b
- a/- b= a/b
This means that the cartridge reaches the maximum position after 3532 seconds. To find the maximum height, we can substitute this value in the expression for the height.
t= 35/32
Calculate power
Multiply
Add fractions
Write fraction as a mixed number
Add terms
Use a calculator
Round to 2 decimal place(s)
This confirms what we see on the graph. The highest point of the path of the cartridge is below the window. Darnell will have 0 chance to catch it.
To find this axis intercept, we need to find the solution of h(t)=0, which is a quadratic equation.
- 16t^2+35t+5=0
Substitute values
Calculate power
Multiply
a(- b)=- a * b
a-(- b)=a+b
Add terms
Now we use a calculator to find approximate values of these solutions. -35+sqrt(1545)/-32&≈ -0.13 -35-sqrt(1545)/-32&≈ 2.32 Since the solution we are looking for represents time, we need the positive solution. The cartridge will hit the ground approximately 2.32 seconds after Jack tossed it.