Kinematics question needed to be solved. Help please

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A basketball is being thrown from a height of 1.5 meters toward a hoop 3.0 meters high, located 4.0 meters away, with an initial velocity of 3.0 m/s. The problem involves determining the angle of launch, using kinematic equations that account for horizontal and vertical motion. A quadratic equation has been derived involving tangent and cosine functions, but further progress has stalled. Suggestions include using a graphing calculator to solve the equation, focusing on angles between 0° and 90°. The discussion emphasizes the importance of an angled throw to avoid hitting the hoop directly.
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Homework Statement



A basketball is being thrown at the hoop (3.0m high) from 4.0 meters away. If the initial height of the ball is at 1.5m with an initial velocity of 3.0 m/s, at what angle should the ball be thrown? (Assume that air resistance is zero, and gravity = 9.81 m/s^2)

Homework Equations



This is a kinematics problem. Even my teacher has trouble solving it. Please help. The main formulas are:

xf = x0 + Vx t + 1/2 a t^2

yf = y0 + Vy t + 1/2 a t^2

The Attempt at a Solution



By using the above two formulas, we came up with a quadratic equation involving tan() and cos() functions. We can't go further. Anyone suggestions?

12 tan(x) - 8.72(1/cos(x))^2 = 1.5

Let x be the starting angle of the ball being being thrown (above the horizontal)

Note: you can't just throw the ball to the hoop in a straight line, you will hit the hoop instead.
 
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if you show how you got the equation someone can help you.
 
Welcome to Physics Forums.

xrayxman said:

Homework Statement



A basketball is being thrown at the hoop (3.0m high) from 4.0 meters away. If the initial height of the ball is at 1.5m with an initial velocity of 3.0 m/s, at what angle should the ball be thrown? (Assume that air resistance is zero, and gravity = 9.81 m/s^2)



Homework Equations



This is a kinematics problem. Even my teacher has trouble solving it. Please help. The main formulas are:

xf = x0 + Vx t + 1/2 a t^2

yf = y0 + Vy t + 1/2 a t^2
Okay, but a would be zero in the xf equation.

The Attempt at a Solution



By using the above two formulas, we came up with a quadratic equation involving tan() and cos() functions. We can't go further. Anyone suggestions?

12 tan(x) - 8.72(1/cos(x))^2 = 1.5
I get almost the same thing, except that I have 4tan(x) instead of 12tan(x).

Let x be the starting angle of the ball being being thrown (above the horizontal)

Note: you can't just throw the ball to the hoop in a straight line, you will hit the hoop instead.
It's probably easiest to use a graphing calculator to solve the equation at this point. You only need to consider x between 0° and 90°, of course.
 
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