Can a Baseball Thrown at 25 m/s Hit a Window 13 Meters High and 50 Meters Away?

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Homework Help Overview

The problem involves a baseball being thrown at a speed of 25 m/s towards a window that is 50 meters away horizontally and 13 meters high. The original poster is trying to determine if the baseball can hit the window and what the maximum height it can reach is, given the constraints of the throw.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the angle of projection and the maximum height achievable. There is a consideration of the optimal angle for maximum range and its implications for hitting a target at a specific height. Some participants suggest treating horizontal and vertical components of motion separately and question the necessity of a 45-degree angle in this context.

Discussion Status

The discussion is ongoing, with participants exploring different angles and their effects on the projectile's trajectory. Some guidance has been offered regarding the separation of horizontal and vertical velocities, and the need to derive equations based on time to find the angle that maximizes height at a given distance.

Contextual Notes

There is an assumption that air resistance can be ignored, and the problem is framed within the constraints of a fixed horizontal distance to the window. The participants are also questioning the appropriateness of the 45-degree angle for this specific scenario.

descartes75
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Hi,
There has to be a simple way of solving this and I am missing it:

A student throws a baseball at a maximum speed of 25 m/s towards a window 50 meter away (horizontally) and 13 meters high. Can he hit the window? (assume the base of the window starts at 13 meters off the ground).
What is the maximum height he can reach from this distance??


Since the Angle is now known and depends on the maximum height, I am having trouble solving this. Any help?
 
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The angle is may not be known but it may be useful to know that 45 degrees is the angle that will give a projectile maximum range in this case, so if the ball can not make it at 45 degrees, then it can not make it period.
 
treat horizontal velocity and vertical velocity separetely,
the horizontal velocity is: 25 * cos(throw_angle)
the vertical velocity is: 25 * sin(throw_angle)

you will want to ignore air resistance, so the horizotal velocity does not change but the vertical velocity does change because of gravity, per time unit it will change 1 g.
so the horizontal velocity will be:
25 * cos(angle)
and the vertical velocity will be:
25 * sin(angle) - gt

can you see why an angle of 45 degrees gets the ball the furthest distance?
 
Thanks Guys.
But isn't 45 degree for maximum range (HOrizontal distance) ??
In this case the range is fixed (50) so intuitively the angle doesn't have to be 45 degrees.

You can imagine standing in front of a tall building with a ball and aiming for a high window. You won't ignore intuition and throw at a 45 degree angle (and hit the door).
 
descartes75 said:
Since the Angle is now known and depends on the maximum height, I am having trouble solving this. Any help?
You'll have to figure out what initial angle gives the maximum height at X = 50m. Start by writing equations for X and Y as a function of time. Then write an expression for Y at the moment that X = 50m. It will be a function of [itex]\theta[/itex]. Find the value of [itex]\theta[/itex] that maximizes this function. (Take a derivative.)
 

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