How Do You Calculate the Required Throwing Speed in Projectile Motion?

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Homework Help Overview

The discussion revolves around calculating the required throwing speed for a quarterback to successfully throw a football to a receiver at a distance of 30 meters, at an angle of 30 degrees. The problem is situated within the context of projectile motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the separation of x and y components of motion, with some questioning the relationship between horizontal distance and maximum height. There are attempts to apply kinematic equations to understand the vertical motion and the implications of initial and final velocities.

Discussion Status

The conversation is ongoing, with participants exploring various aspects of projectile motion. Some have provided insights into the behavior of vertical velocity at maximum height, while others are still seeking clarity on how these concepts relate to finding the initial velocity needed for the throw.

Contextual Notes

Participants express uncertainty regarding the application of kinematic equations and the assumptions about the height of release and reception being the same. There is also a mention of needing to clarify the relationship between the components of velocity and the overall motion of the projectile.

P(physics)DD
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Need help on projectile motion equation!

okay today i dazed off in class and didn't really listen to how my teach told me how to do projectile motion equations. here's the one I'm working on now.

A quaterback tries to throw a football to his receiver at an angle of 30 degrees with respect to the ground. The receiver is at a distance 30 meters from the quarterback.

(a) The receiver stands still. At what speed must the quarterback throw the ball so the receiver can catch it? (Assume that the paint of release and the point of reception are at the same height)

i've already split the x and y components apart and wrote down what i know. other than that I am seriously stuck.

Would draw a picture for you but can't really :) please somebody get back to me
 
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What do you know about the distance (horizontally) that the ball has traveled when it reaches its maximum height?
 
idk if i were to guess it'd be half the x component wouldn't it
 
P(physics)DD said:
idk if i were to guess it'd be half the x component wouldn't it
Vertically, the speed would be Vsin30, right?
since acceleration=g which is constant, the kinematics equations apply

so if [itex]v=u+2at[/itex] where v=final vertical speed,u=initial vertical speed


When the ball reaches the maximum height, what would be its final velocity?(Hint:As the ball goes up, the final velocity decreases)
 
why would the vertical be vsin30?
 
The speed at 30 degrees is V. When you spit that into the x and y components, the initial vertical velocity is Vsin30
 
oh okay that makes sense now.
working it out now
 
rock.freak667 said:
Vertically, the speed would be Vsin30, right?
since acceleration=g which is constant, the kinematics equations apply

so if [itex]v=u+2at[/itex] where v=final vertical speed,u=initial vertical speed


When the ball reaches the maximum height, what would be its final velocity?(Hint:As the ball goes up, the final velocity decreases)

how does this help me find the initial velocity of the ball
 
The Vsin30 is the initial velocity.At the maximum height what do you think the final velocity is?
 
Last edited:
  • #10
idk...you got to kinda spoon feed me here. lol
 
  • #11
Well as the ball rises..the vertical velocity decreases...until it reaches the max height. After reaching the maximum height, the vertical velocity begins to increase...

the only for that to happen is if the velocity at the max height is Zero.

and the time taken to reach the maximum height is the same as the time taken to go from the max height to the ground.

(Note: this is just one method, another kinematic equation can be used to find the time taken to reach the receiver as well)
 
  • #12
have u got the answer because I've gotten 26.1 m/s
 

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