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

In summary, the student can hit the window if he throws the ball at a 45 degree angle and if he throws it at a maximum speed of 25 m/s.
  • #1
descartes75
3
0
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 hight he can reach from this distance??


Since the Angle is now known and depends on the maximum hight, I am having trouble solving this. Any help?
 
Physics news on Phys.org
  • #2
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.
 
  • #3
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?
 
  • #4
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).
 
  • #5
descartes75 said:
Since the Angle is now known and depends on the maximum hight, 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.)
 

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

What is "Stuck on Simple Projectile Motion"?

"Stuck on Simple Projectile Motion" is a phrase used to describe a common problem that students encounter when learning about projectile motion in physics. It refers to the difficulty of solving problems involving the motion of objects launched or thrown into the air.

Why do students often struggle with "Stuck on Simple Projectile Motion"?

Students may struggle with "Stuck on Simple Projectile Motion" because it requires a combination of mathematical skills and understanding of physical concepts. It also involves visualizing the motion of an object in two dimensions, which can be challenging for some students.

What are the key concepts to understand in order to solve problems related to "Stuck on Simple Projectile Motion"?

The key concepts to understand in order to solve problems related to "Stuck on Simple Projectile Motion" include the laws of motion, gravity, and the concept of velocity and acceleration in two dimensions. It is also important to understand how to break down a projectile motion problem into its horizontal and vertical components.

What are some tips for solving problems related to "Stuck on Simple Projectile Motion"?

Some tips for solving problems related to "Stuck on Simple Projectile Motion" include drawing a diagram to visualize the motion of the object, breaking the problem down into its horizontal and vertical components, and using equations such as the kinematic equations to solve for the unknown variables.

How can students improve their understanding of "Stuck on Simple Projectile Motion"?

Students can improve their understanding of "Stuck on Simple Projectile Motion" by practicing problems and seeking help from their teacher or peers if they are struggling. They can also use online resources and simulations to better understand the concepts and see them in action.

Back
Top