Throwing a Ball: Calculating Velocity & Flight Time

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The discussion focuses on calculating various aspects of a ball thrown at an angle, reaching a height of 8.0m and traveling 16m horizontally. The vertical velocity component at release can be determined using the formula v_{0y}=\sqrt{2gH}, where g is the acceleration due to gravity. The total flight time is calculated as Δt=2t, with t derived from the vertical velocity. The horizontal velocity at release is found using v_{0x}=\frac{L}{\Delta t}. Finally, the overall speed at the release point is calculated using the Pythagorean theorem, v_0=\sqrt{v_{0x}^2+v_{0y}^2}.
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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.hight 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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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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