An arrow has an initial launch speed of 39 m/s. If it must strike a target 39 m away at the same elevation, what should be the projection angle?(Hint: sin2θ = 2sinθcosθ)
u = 39 cos T
39 = u t = 39 cos T * t
t = 1/cosT total time in air
rise time = t/2 = 1/(2cosT)
vertical problem
0 = 39 sin T - g * 1/(2 cos T)
but g/2 = 4.9 m/s^2
4.9/cos T = 39 sin T
cos T sin T = 4.9/39
2 sin T cos T = 9.8/39
so
sin 2T = .251
2T = 15.55
T = 7.28 degrees
To find the projection angle of the arrow, we can use the following equation:
Range = (v^2 * sin(2θ)) / g
Where:
- Range is the horizontal distance traveled by the arrow (given as 39 m)
- v is the initial launch speed of the arrow (given as 39 m/s)
- θ is the projection angle we need to find
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Let's plug in the values we know and solve for θ:
39 = (39^2 * sin(2θ)) / 9.8
To simplify the equation, we can use the hint provided:
sin(2θ) = 2sin(θ)cos(θ)
39 = (39^2 * 2sin(θ)cos(θ)) / 9.8
Now, let's rearrange the equation:
sin(θ)cos(θ) = (39 * 9.8) / (2 * 39^2)
sin(θ)cos(θ) = 0.05
We can recognize that sin(θ)cos(θ) represents the sine of twice the angle, so we can rewrite:
sin(2θ) = 0.05
Now, we can solve for 2θ:
2θ = arcsin(0.05)
θ = (1/2) * arcsin(0.05)
Using a scientific calculator or tool to find the arcsin of 0.05, we get:
θ ≈ 5.7 degrees
Therefore, the projection angle of the arrow should be approximately 5.7 degrees.