What is the Time of Collision for Two Vertically Launched Objects?

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The discussion revolves around the collision of two vertically launched arrows, with the first arrow shot upward at 25 m/s and the second arrow launched halfway to the peak. The first arrow reaches a maximum height of approximately 31.86 meters, requiring a specific time to reach that peak. The second arrow's launch speed and maximum height are calculated based on the timing of the collision, which occurs when the first arrow is at its highest point. Additionally, a scenario involving two stones dropped and thrown from a cliff is presented, requiring calculations for the time of their meeting. The thread emphasizes the importance of understanding projectile motion and the equations of motion to solve these problems effectively.
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Pat shoots an arrow vertically upward with an initial speed of 25 m/s. When it is exactly halfway to the peak of its flight, he shoots a second arrow vertically upward from the same spot. They collide just as the first arrow reaches its highest point.

(A) What is the Launch Speed of the Second Arrow?
(B) What Maximum Height does the second arrow Reach?


Brad is on Top of a 20m Cliff. Below he spots Tom. He drops a stone off the cliff. At the same time Tom throws a second stone vertically upwards with an initial velocity of 30 m/s. Assuming that he threw the rock hard enough, at what time will the two stones meet?
 
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Givens:
Via1 = 25 m/s
a = -9.81m/s^2
Vfa1 = 0 m/s
Via2 = ?
da1 = ?

Vf^2 = Vi^2 + 2ada1
0^2 = 25^2 + 2(-9.81)da1
-625 = -19.62da1
da1 = 31.85524 m

Therefore Half the height = 15.92762 m

Need Help Continuing

 
Okay now compute the times...How much time will the second arrow require to reack the peak and how is that related with the time the first arrow travels the same distance?

Daniel.
 

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