Projectile Motion at an Angle: Calculating Time and Distance

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Homework Help Overview

The problem involves projectile motion, specifically calculating the time it takes for a seed to land after being shot from a seed pod at an angle below the horizontal. The initial velocity and height from which the seed is launched are provided.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for vertical and horizontal components of the initial velocity, as well as the application of kinematic equations to determine time and distance. There is an attempt to solve a quadratic equation derived from the motion equations.

Discussion Status

Some participants have provided calculations and attempted to solve for time using the quadratic formula. There is an acknowledgment of a potential error in the initial solution provided by the original poster, prompting further examination of the calculations.

Contextual Notes

Participants are working under the constraints of the problem as stated, including the initial velocity, angle, and height. There is a focus on ensuring that the signs of the variables are consistent with the direction of motion.

MozAngeles
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Homework Statement


When the dried-up seed pod of a scotch broom plant bursts open, it shoots out a seed with an initial velocity of 2.64 m/s at an angle of 30.0˚ below the horizontal. The seed pod is 0.465 m above the ground.
How long does it take for the seed to land?
What horizontal distance does it cover during its flight?


Homework Equations


Voy=Vsin\theta
Vox=Vcos\theta
h = y0 + v0*t + 1/2 * a * t^2
d= Vox*t

The Attempt at a Solution


I used Voy=2.64*sin30˚, Vox= 2.64cos30 I plugged everything into these equations, solved for t using the quadratic equation, and I got t=0.4723 but that was wrong
 
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Show your calculations.
 
Voy=2.64sin30= 1.33

h = y0 + v0*t + 1/2 * a * t^2
-0.465= 1.33t + 1/2*(-9.8)t^2
-4.9t^2 + 1.33t+ .465=0
t= -1.33-sqrt 1.33^2-(4*-4.9*.465)/ 2*-4.9
t= .04723
 
MozAngeles said:
Voy=2.64sin30= 1.33

h = y0 + v0*t + 1/2 * a * t^2
-0.465= 1.33t + 1/2*(-9.8)t^2
-4.9t^2 + 1.33t+ .465=0
t= -1.33-sqrt 1.33^2-(4*-4.9*.465)/ 2*-4.9
t= .04723

vo, h and g are in the same direction. So they must have the same sign.
 
Thank you so much, that helped!
 

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