What is the maximum distance a stone thrown at 30 m/s can cross a river?

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Homework Help Overview

The problem involves determining the maximum distance a stone can be thrown across a river when thrown at an initial speed of 30 m/s, with the condition that it lands at the same vertical height from which it was thrown. The discussion revolves around projectile motion concepts.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to calculate time of flight and horizontal distance using projectile motion equations. There are attempts to derive the range equation and questions about the correct angle of launch for maximum distance. Some participants express confusion over their calculations and results.

Discussion Status

The discussion is ongoing, with various participants sharing their calculations and questioning each other's results. Some guidance has been offered regarding the use of trigonometric components of the initial velocity, but no consensus has been reached on the correct approach or final answer.

Contextual Notes

Participants are working under the assumption that the stone lands at the same height from which it was thrown, and there are indications of differing interpretations of the time of flight and horizontal distance calculations.

roxxyroxx
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Homework Statement



A stone is thrown at 30 m/s. what is the widest river it can be thrown across? (the stones ends at the same vertical height from where it was thrown)

Homework Equations



x=vxot
y=vyot + 0.5gt2

The Attempt at a Solution


i know i need to find t first. using the second equation i got t as 5.2s. when i multiplied 5.2s by vxo (30m/s) i got 156.3 m but the answer should be 92m. wat did i do wrong?
 
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You can't just use the V as Vx.

Vo is constrained to be 30 m/s.

I rather think they want you to derive the Range Equation and determine where range reaches its maximum.

Using your equations then, and

Vox = Vo*Cosθ

and

Voy = Vo*Sinθ

derive an expression for X, total distance, and then determine at what angle of launch you get the maximum.

Along the way you might find this useful ... 2*Cosθ*Sinθ = sin(2θ)
 
well when i used x=vxot
i put in (30cos45)(5.2s) but i got 82m instead of 92 m
 
roxxyroxx said:
well when i used x=vxot
i put in (30cos45)(5.2s) but i got 82m instead of 92 m

You maybe want to recalculate.

I get a different time which is less and the answer given - after rounding.
 
how did u get ur time?
i can't seem to get any other answer than what i got..
 
v = g*t

t = v/g

That's time to max. Double it to come back down.

v = 30*sin45 = 30*√2/2

30*√2/2*2/g = 30*√2/g = ... ?
 
Last edited:

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