Projectile motion apple throw problem

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  • #51
gneill
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Gneill when i use the quadratic form, and plug in the given variables i do not get 1.27 or 1.23 or anywhere near that
∆t = [-2.86+/- sqrt( -2.86^2-4(4.906)(-4)]/(2(4.906) )
t=-3.8 or -1.91

The 'b2' term inside the square root should not be negative. The 'c' term is not negative in this case, it's the initial height which is positive.

Initial height is +4m. Final height is 0.00m (ground).

The kinematic equation in the vertical direction is

[itex] 0 = h + v_y t - \frac{1}{2}g t^2 [/itex]
 
  • #52
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thanx a million i know see were i went wrong yup its equal to 1.23 sec
for b i used dx=(Vx)(t)
Vx=5.0 m/s x Cos35=4.09m/s
dx=(4.09m/s)(1.23s)
dx= 5.03m
for c i did V^2=(5.0m/s)^2+2g(4m)
V= 10.16 m/s
i just cant figure out how to get the angle,i know im supposed to use the tan function to find the angle i plug in 10.16*tan(35)=7.14 hmm this does not make sense? were am i going wrong?
 
  • #53
gneill
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You have found the speed of the apple when it reaches the ground (that's its speed along its direction of motion). To find the angle that this speed makes with respect to some reference direction (which direction do you think would be appropriate?) you should draw a picture showing this speed along the hypotenuse of a triangle along with another component that you know as one of the sides. Solve for whatever angle is appropriate.
 

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