Projectile Motion: Calculating Initial Velocity for Grenade Throw in Action Film

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SUMMARY

The discussion focuses on calculating the initial velocity required for a grenade throw in an action film scenario where the hero's car travels at 90.0 km/h and the enemy's car at 127 km/h, with a distance of 15.6 m between them. The grenade is thrown at a 45-degree angle relative to the hero's position. To solve for the initial velocity, one must determine both the horizontal and vertical components of the throw, factoring in the relative speeds of the cars and the time taken for the grenade to travel the distance. The total distance equation is given as X final = 15.6 + 37t, where 't' represents time.

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limekiwi
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In an action-adventure film, the hero is supposed to throw a grenade from his car, which is going 90.0 km/h, to his enemy's car, which is going 127 km/h. The enemy's car is 15.6 m in front of the hero's when he let's go of the grenade.
If the hero throws the grenade so its initial velocity relative to him is at an angle of 45 degrees above the horizontal, what should the magnitude of the initial velocity be? The cars are both traveling in the same direction on a level road. You can ignore air resistance.

I've bee reading this problem over and over again, and don't really know how to start. Since I am supposed to find initial velocity, I need the vertical and horizontal component. But from the question, I don't know how to find the horizontal component. Can someone please lead me in the right direction?
 
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This question didnt make a lot of sense to me... I don't see how it can be solved numerically. The total distance the grenade needs to travel, X final = 15.6 + 37t

to find the velocity, you need the total distance, to get the the total distance you need to know the time it took for the grenade to get from one car to the other. Then again, I'm no rocket scientist, maybe I'm misunderstanding the problem. *shrugs shoulders*
 

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