Arrow Trajectory: 29° and 50 m/s

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An arrow is shot at a 29° angle with an initial velocity of 50 m/s, and the discussion focuses on calculating its maximum height and horizontal distance to the target. To find the maximum height, the vertical component of the velocity should be analyzed using the equation Vf^2 = Vi^2 + 2ad. The time the arrow remains in the air can be determined first, which is then used to calculate the horizontal distance by multiplying this time by the horizontal component of the velocity. The emphasis is on starting calculations with the vertical component for gravity-related problems. Understanding these steps is crucial for solving projectile motion scenarios effectively.
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An arrow is shot at 29.0° above the horizontal. Its velocity is 50 m/s and it hits the target.

What is the maximum height the arrow will attain?


The target is at the height from which the arrow was shot. How far away is it?

if anyone could give me some tips or steps to solve this problem that would be great

thanks
 
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bam3211 said:
An arrow is shot at 29.0° above the horizontal. Its velocity is 50 m/s and it hits the target.

What is the maximum height the arrow will attain?
Vf^2 = Vi^2 + 2ad



The target is at the height from which the arrow was shot. How far away is it?
How long does the arrow stay in the air? Find that time then multiply it by the horizontal component of the arrow's velocity.


For most (all?) of these gravity related problems, your work should start with the vertical component.
 
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