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

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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 = ... ?
 
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