Throwing a Short Put - simple Distance problem?

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SUMMARY

The discussion centers on calculating the new range of a shotput thrown at a 55-degree angle after previously achieving a distance of 30 meters at a 45-degree angle. The user expresses confusion regarding the application of the projectile motion equations, specifically the formula for distance, which includes components of acceleration and velocity. The correct approach involves using the range formula for projectile motion, which is given by R = (v^2 * sin(2θ)) / g, where v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. The calculated new range at 55 degrees is approximately 41 meters.

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Homework Statement


Boy throws a shotput at angle of 45Degree which travels a horizontal distance of 30m
If he releases same shotput at same speed but changes angle to 55 Degree, what will be the new range or distance?
A) 25m B)41m C)28m D)37m E)32. m

Homework Equations


That's what i am finding.?/?


The Attempt at a Solution



I tried using the info posted somewhere like
Distance = 1/2 * (acceleration) * (time squared) + (Velocity) * (Initial time) + (Initial Distance)
But i am not able to get what to put where? what to put in velocity? horizontal or vertical?
I am not even sure this is correct way.
Pleas hlp...driving me nuts!
Just explain me the concept with equation, if you do not want to solve for me. please
 
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Thanks for your help. That worked nicely :biggrin:
 

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