Calculate Time Until Tree is Hit: 9m Away, 18m/s Velocity

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SUMMARY

The problem involves calculating the time it takes for a ball thrown horizontally at 18 m/s from a height of 1.5 m to hit a tree located 9 m away. The horizontal velocity remains constant, and the vertical motion is influenced by gravity. The time to hit the tree can be calculated using the formula \( t = \frac{d}{v} \), where \( d \) is the horizontal distance (9 m) and \( v \) is the horizontal velocity (18 m/s), resulting in a time of 0.5 seconds. The height at which the ball hits the tree can be determined using the vertical motion equations, factoring in the time of flight and gravitational acceleration.

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Homework Statement



- ball is thrown with a horizontal velocity of 18 m/s directly toward a tree
- person who threw the ball is 1.5 m above the ground and 9 m away from the tree

looking for time it takes to hit the tree, what height the ball hits the tree at, and the balls velocity when it does so.

Homework Equations



Well, we know the velocity of x which is 18 m/s, the initial y velocity which is zero, the distance between the ball and the tree which is 9m and the initial height of the ball which 1.5m.

To find time there are many formulas but they all have (delta)d of y, but u don't have d2 so i can't find it using any of those. After getting time i can the formulas to find the others.

The Attempt at a Solution

No idea. No formulas that have what i need.

Do i need to first find the y final velocity using a different formula??
 
Last edited:
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i would just use v=x/t to solve for time because there is no acceleration.
 
Thanks.

I forgot that vx stays constant.
 

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