Answer is 3 seconds
When the bullet reaches the ground, ground being x in graph (and here its s which is = 0)
s = -16t^2 + 48t
s = 0, solve for t
0 = -16t^2 + 48t
0 = t ( -16t + 48)
0 = 16t ( - t + 3)
now you have two equation
0 = 16t and 0 = -t +3 ( you can look at the graph line touches x twice)
0 = 16 t
0 = t ( you know its false, because time = 0)
You are left with
0 = -t + 3
t = 3
It takes 3 seconds for the bullet to return to the ground.
// Hope this helps.
Answer:
Step-by-step explanation:
Straightforward use of the inverse tangent function of a calculator will tell you θ ≈ -0.08678 radians. This is an angle in the 4th quadrant, where your restriction on θ places it. (To comply with the restriction, you would need to consider the angle value to be 2π-0.08678 radians. The trig values for this angle are the same as the trig values for -0.08678 radians.)
Likewise, straightforward use of the calculator to find the other function values gives ...
sin(-0.08678 radians) ≈ -0.08667
cos(-0.08678 radians) ≈ 0.99624
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<em>Note on inverse tangent</em>
Depending on the mode setting of your calculator, the arctan or tan⁻¹ function may give you a value in degrees, not radians. That doesn't matter for this problem. sin(arctan(-0.087)) is the same whether the angle is degrees or radians, as long as you don't change the mode in the middle of the computation.
We have shown radians in the above answer because the restriction on the angle is written in terms of radians.
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<em>Alternate solution</em>
The relationship between tan and sin and cos in the 4th quadrant is ...
That is, the cosine is positive, and the sign of the sine matches that of the tangent.
This more complicated computation gives the same result as above.