Projectile Motion: Calculations for a successful basketball shot

AI Thread Summary
The discussion centers on the assumption that a basketball reaches its maximum height at the midpoint of its trajectory. Participants question the validity of this assumption and suggest that a clearer diagram could enhance understanding. They emphasize the importance of providing a rationale for such assumptions and encourage exploring alternative trajectories. The conversation highlights the need for precise calculations regarding horizontal and vertical displacements as the ball approaches the hoop. Ultimately, clarity in the calculations and assumptions is deemed essential for successfully analyzing projectile motion in basketball shots.
ayderay
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Homework Statement
A basketball player shoots successfully at goal from a horizontal distance of 5.3m to the centre of the goal ring. She releases the ball at an angle of 48 to the horizontal and 1.2m below the height of the ring. What was the ball’s speed as it left her hand? The answer is 8.1m/s
Relevant Equations
Motion equations
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Why have you assumed that the ball reaches the maximal height at the midpoint of the trajectory? Can you make an argument for this?
 
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Orodruin said:
Why have you assumed that the ball reaches the maximal height at the midpoint of the trajectory? Can you make an argument for this?
I assumed it..
 
Perhaps a clearer diagram might help. We can ignore the practicalities of the hoop, and just consider the target as a point in space that we want to reach, at the top of the pole:

1587288195354.png


There is another trajectory you could have, I'll let you think of what that might be! It won't affect your calculations.
 
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ayderay said:
I assumed it..
It is not a healthy habit to assume things that you cannot give an argument for. Can you show whether it is the case or not?
 
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As the ball reaches the ring, how much is its horizontal displacement? How much is its vertical displacement?
 
etotheipi said:
Perhaps a clearer diagram might help. We can ignore the practicalities of the hoop, and just consider the target as a point in space that we want to reach, at the top of the pole:

View attachment 260947

There is another trajectory you could have, I'll let you think of what that might be! It won't affect your calculations.
Ohh, thank u for this, it’s much more understandable. I’ll try the question out again
 
ayderay said:
Ohh, thank u for this, it’s much more understandable. I’ll try the question out again
You treat 1.2m as displacement along vertical.
5.3 m as displacement along horizontal.
And proceed. You will get
 
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