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Answer :
Answer:
C. 28.2 deg
Explanation:
The horizontal range of a projectile is given as:
[tex]R = \frac{v^2Sin2\theta}{g}[/tex]
where,
R = Range
v = speed
θ = angle of launch
g = acceleration due to gravity = 9.81 m/s²
First, we will find the launch speed (v) by using the initial conditions:
R = 120 m
θ = 45°
Therefore,
[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]
Now, consider the second scenario to hit the target:
R = 100 m
Therefore,
[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]
Hence, the correct option is:
C. 28.2 deg
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Final answer:
To hit the target, the cannon can be adjusted to angles of 49.1 degrees and 52.8 degrees.
Explanation:
To find the possible angles that the cannon can be adjusted to hit the target, we need to use the concept of projectile motion. Since the cannonball overshoots the target, we know that the initial velocity of the cannonball has a vertical component and a horizontal component. Using trigonometry and the given information, we can determine the initial velocity of the cannonball and then find the possible angles. In this case, the cannon can be adjusted to angles of 49.1 degrees and 52.8 degrees to hit the target.
Learn more about Projectile motion here:
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