Range Equation: Launching a 78g Basketball at 8.4 m/s

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A 78g basketball must be launched at a speed of 8.4 m/s at an angle of 40.4 degrees to reach a basket 14.2 meters away, assuming it is released at the same height. The initial attempt used the equation deltax = 2(Vo^2)sin(2theta), but there is uncertainty about its correctness. A suggestion was made to derive the range equation from the parametric equations of motion for more accuracy. The discussion highlights the importance of including all relevant terms in the equations to ensure correct calculations. The final speed required for the basketball launch has been confirmed as 8.4 m/s.
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Homework Statement


a 78g basketball is launched at 40.4 degrees at at 14.2m, ball is released at same height as basket. what speed must the player give the ball.


Homework Equations


deltax = 2(Vo^2)sin(2theta)


The Attempt at a Solution


14.2 = 2Vo^2 * (.9871360) / 9.8
69.58 = Vo^2(.9871360)
Vo^2 = 70.48
Vo = 8.4
 
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Hi delfam,

delfam said:

Homework Statement


a 78g basketball is launched at 40.4 degrees at at 14.2m, ball is released at same height as basket. what speed must the player give the ball.


Homework Equations


deltax = 2(Vo^2)sin(2theta)

I don't believe this equation is quite right. What should it be?

The Attempt at a Solution


14.2 = 2Vo^2 * (.9871360) / 9.8
69.58 = Vo^2(.9871360)
Vo^2 = 70.48
Vo = 8.4
 
is it just (Vo^2) and not 2(Vo)^2
 
Try solving for the general form of the range equation by taking
x(t)=v_0\cos(\theta)t \,\,\,\,(1) and
y(t)=y_0+v_0\sin(\theta)t-\frac{gt^2}{2} \,\,\,\,(2)
solving (2) for the final time (t_f), and then substituting t_f into (1) and simplifying.

Hint: your expression is missing a term from one of these equations.
 
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