Projectile Motion at an Angle: Calculating Time and Distance

AI Thread Summary
The discussion focuses on calculating the time and horizontal distance a seed travels when launched from a scotch broom plant at an angle of 30 degrees below the horizontal with an initial velocity of 2.64 m/s from a height of 0.465 m. The participant attempted to solve for time using the equations of motion but initially arrived at an incorrect time of 0.4723 seconds. They correctly calculated the vertical component of velocity (1.33 m/s) and set up the quadratic equation for height, leading to a revised time calculation. The importance of ensuring that the signs of initial velocity, height, and gravity are consistent in the equations was emphasized. The discussion concludes with appreciation for the clarification provided.
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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