- #1
cj
- 85
- 0
My Baseball coach is also my physics instructor. He says that "it is easier to throw a baseball horizontally than it is vertically" -- and that throwing speed varies with elevation angle approximately as:
[tex]v_0cos(\frac {\theta}{2}) \text{ m/s}[/tex]
[tex]\text {where }v_0 = \text{the initial velocity when the ball is thrown horizontally, and}[/tex]
[tex] \theta_0 =\text { the initial elevation angle}[/tex]
[tex] \text{At what }\theta_0 \text{ do I need to throw the ball to achieve maximum height and, conversely, range?}[/tex]
I have a feeling this is a "classic" physics problem. I can't, however, think of the velocity versus launch angle conditions leading to maximum height or range.
The only facts I can cough-up is that at maximum height, vertical velocity = 0. Also, maximum range occurs when the baseball experiences maximum flight time along with maximum horizontal velocity.
Using these relationships to solve for optimum initial launch angle is stumping me.
[tex]v_0cos(\frac {\theta}{2}) \text{ m/s}[/tex]
[tex]\text {where }v_0 = \text{the initial velocity when the ball is thrown horizontally, and}[/tex]
[tex] \theta_0 =\text { the initial elevation angle}[/tex]
[tex] \text{At what }\theta_0 \text{ do I need to throw the ball to achieve maximum height and, conversely, range?}[/tex]
I have a feeling this is a "classic" physics problem. I can't, however, think of the velocity versus launch angle conditions leading to maximum height or range.
The only facts I can cough-up is that at maximum height, vertical velocity = 0. Also, maximum range occurs when the baseball experiences maximum flight time along with maximum horizontal velocity.
Using these relationships to solve for optimum initial launch angle is stumping me.