Throwing a Ball: Calculating Velocity & Flight Time

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SUMMARY

The discussion focuses on calculating the velocity components and flight time of a ball thrown at an angle. The vertical height reached by the ball is 8.0m, and it travels a horizontal distance of 16m. The vertical velocity component at release is calculated using the formula \(v_{0y}=\sqrt{2gH}\), where \(g\) is the acceleration due to gravity. The total flight time is determined to be \(\Delta t=\frac{v_{0y}}{g}\) and the horizontal velocity component is found using \(v_{0x}=\frac{L}{\Delta t}\).

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rachael
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A boy throws a ball into the air at an angle to the horizontal. The ball reaches a vertical height of 8.0m and travel a horizontal distance of 16m before being at the same vertical position as at the point of release
A. what is the magnitude of the vertical velocity component of the ball as it is thrown?
B. What is the time of the flight of the ball?
C. What is the horizontal velocity component of the ball at its point of throwing?
d. What is the speed of the ball at this point of release?
Thanks
 
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Find the time to fall down from the top first. the time to rise is the same.
That's (B).
Then do (c) or (A).
 
rachael said:
A boy throws a ball into the air at an angle to the horizontal. The ball reaches a vertical height of 8.0m and travel a horizontal distance of 16m before being at the same vertical position as at the point of release
A. what is the magnitude of the vertical velocity component of the ball as it is thrown?
B. What is the time of the flight of the ball?
C. What is the horizontal velocity component of the ball at its point of throwing?
d. What is the speed of the ball at this point of release?
Thanks

A) The maximum height of the ball is H=\frac{v_{0y}^2}{2g}
\Longrightarrow v_{0y}=\sqrt{2gH}

B)The vertical component of the ball v_y=v_{0y}-gt

The ball reach the max.height when v_y=0\Longrightarrow t=\frac{v_{0y}}{g}

The time of the flight of tha ball is \Delta t=2t=\frac{v_{oy}}{g}

C) The horizontal velocity component of the ball at its throwing point is v_{0x}=\frac{L}{\Delta t}

D) The speed of the ball at release point is v_0=\sqrt{v_{0x}^2+v_{0y}^2}[/color]
 
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