Projectiles launched horizontally

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The discussion centers around calculating the speed of a baseball pitched horizontally, using the equation for vertical displacement under gravity. The key variables are identified: vertical displacement (y), acceleration due to gravity (g), and time of flight (Δt). Participants express confusion about how to incorporate horizontal displacement into the calculations. The solution involves first determining the time of flight using the vertical displacement equation, then using that time to calculate the initial horizontal velocity. Clarification is sought on isolating the time variable and the correct application of the formula.
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okay so here's the problem..

the fastest recored pitch in MLB was thrown by nolan ryan in 1974. if this pitch were thrown horizontally, the ball would fall 0.809 m by the time it reached the home plate, 18.3 m away. how fast was this pitch?

the book suggests the equation
delta y = -1/2g(deltat)squared

we just started this lesson and I'm really not sure what variable represents what in this problem :confused:
 
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If you simply want to know how to use the formula, then the variables are as follows;

y is the vertical displacement
g is the acceleration due to gravity
Δt is the time of flight
 
well, would it be .809=-1/2(g)(t)squared, what else are you supposed to do with this? like where does the horizontal displacement go
 
rachelx46 said:
well, would it be .809=-1/2(g)(t)squared, what else are you supposed to do with this? like where does the horizontal displacement go
That's right. You now need to solve for the time of flight t. You can then use this together with the horizontal displacement to find the initial velocity.
 
ahh I am completely lost now i tried to isolate teh unknwon (t) from the rest of the problem and got t= the square root of 2(delta y)/-g

but the book says that is only for when delta y is negative. can someone explian this problem to me step by step? I am reallly confused.
 
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