Projectile Motion Arrow Problem

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SUMMARY

The discussion focuses on solving a projectile motion problem involving an arrow shot at an angle of 31.0° with an initial velocity of 53 m/s. The maximum height attained by the arrow is calculated using the formula Max Height = Vi*T + 1/2*-9.8*T^2, where T is derived from the vertical component of the initial velocity. The time to reach maximum height is determined to be 2.18 seconds, using the equation T = (53*sin(31.0°))/9.8. The horizontal distance to the target, which is at the same height as the launch point, can be calculated using the horizontal velocity component.

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  • Understanding of projectile motion principles
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  • Knowledge of trigonometric functions for angle decomposition
  • Basic physics concepts including gravity and acceleration
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davetheant
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An arrow is shot at 31.0° above the horizontal. Its velocity is 53 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?


It says to neglect air resistance so acceleration in the x direction must be 0. Obviously gravity makes the acceleration in the y direction -9.8

I tried to calculate the maximum height by using the formula Max Height = Vi*T+1/2*-9.8*T^2

I got T to equal 2.18 by [53*sin(40)]/9.8
please help
 
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In the first case, the initial velocity in the upward direction is known, acceleration is given and final velocity at the maximum height is known(?). Find the maximum height. Which kinematic equation relates all these quantities?
 

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