Calculate Max Height & Distance of Arrow Shot at 30°

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SUMMARY

The discussion focuses on calculating the maximum height and horizontal distance of an arrow shot at a 30° angle with an initial speed of 50 m/s. The maximum height is determined using the formula for vertical motion, specifically H = (v^2 * sin²(θ)) / (2 * g), where g is the acceleration due to gravity (9.81 m/s²). The horizontal distance can be calculated using the time of flight and horizontal velocity, with the formula R = (v² * sin(2θ)) / g. The participant initially misapplied the sine function to find the height, indicating a need for clarity on the correct equations for projectile motion.

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  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for constant acceleration
  • Basic physics concepts such as gravity and velocity
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  • Learn how to apply kinematic equations for projectile motion
  • Study the derivation of the maximum height formula for projectiles
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Homework Statement



An arrow is shot at 30.0° above the horizontal. Its initial speed is 50 m/s and it hits the target.

(a) What is the maximum height the arrow will attain?
(b) The target is at the height from which the arrow was shot. How far away is it?



Homework Equations



SOH CAH TOA

O = H sin(theta)
A = H cos(theta)


The Attempt at a Solution



For my attempt at a, I simply drew it out at first. I made a vector line pointing 30 degrees NE and 50m/s long and formed a grid around it so i could plot out a right traingle. With the measurement of the hypotunuse (50) and the degree of the angle (30), I thought I just had to find the other sides to determine the solution. So i did 50sin(30) to get 25 for the y-axis rise, but that was not the answer.

I havnt attempted part b yet because I do not know where to proceed.
 
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help me
 
You need to use the equations for constant acceleration . We are given the initial velocity, so which of those equations might be helpful here?
 

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