Easy kinematic problem I've suffered an hour for

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SUMMARY

The discussion focuses on solving a kinematic problem involving a metal ball launched at a 30-degree angle from an initial height of 1 meter with an initial speed of 7.66 m/s. Participants emphasize the use of the kinematic equation s = ut - 0.5gt² to calculate vertical and horizontal distances. The vertical component of the initial velocity is determined using u = 7.66 * sin(30), while the horizontal component is calculated using u = 7.66 * cos(30). The time to reach the floor is derived from a quadratic equation, leading to the calculation of horizontal displacement as the product of horizontal speed and time.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with kinematic equations
  • Basic trigonometry for resolving velocity components
  • Knowledge of gravitational acceleration (g = 9.8 m/s²)
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  • Study the derivation and application of kinematic equations in projectile motion
  • Learn how to resolve vectors into their components using trigonometric functions
  • Explore the concept of quadratic equations and their solutions in physics problems
  • Practice similar projectile motion problems to reinforce understanding
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Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators seeking to enhance their teaching methods in these topics.

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Homework Statement



Metal ball is launch at 30 degrees from initial height of 1m with initial speed as 7.66m/s, find horizontal distance traveled

Homework Equations





The Attempt at a Solution



Honestly I have no idea.
 
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Go through any textbook or hyper Physics site and collect the information about the projectile motion. If you have any doubts, please feel free to ask us. Before that you have to show your attempt.
 
you have to use the kinematics equation s= u*t - 0.5gt^2 in vertical and horizontal directions. u is the initial velocity and s is the distance travelled. g is the gravi . acc = 9.8 ms-2 , when it reaches floor level it has traveled -1m from the place of launch. use above equation for vertical direction with s= -1m , and u=7.66* sin30 , you have to resolve for the vertical component of launch speed, now when you plug in this data in above equation you will get a quadratic equation in t you can find two roots to this equation , one root will be (-) so you can take the positive root, this is the time it takes to get to the floor level, now think of the horizontal displacement there is no acceleration in that directin so the resolved horizontal speed ( 7.66*cos 30) will remain unchanged so the horizontal dispacement = horizontal speed * time it takes to getto the floor !
good luck
 
rl.bhat said:
Go through any textbook or hyper Physics site and collect the information about the projectile motion. If you have any doubts, please feel free to ask us. Before that you have to show your attempt.
check my previous posts, I am not a leech
 
What's your previous attempts though?

Did you try breaking it down into the x- and y-components?

Even a sit down with a sketch to familiarise yourself with the problem helps.

:smile:
 

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