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

AI Thread Summary
To calculate the required throwing speed for a quarterback to successfully throw a football to a receiver 30 meters away at a 30-degree angle, the initial vertical velocity is determined using Vsin30. The ball's vertical speed decreases until it reaches maximum height, where the final vertical velocity is zero. The time to reach maximum height equals the time to descend back to the same height. Using kinematic equations, the calculated initial speed needed for the throw is approximately 26.1 m/s. Understanding these principles of projectile motion is essential for solving similar problems.
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 v=u+2at 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 v=u+2at 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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