Stuck on Simple Projectile Motion

1. Oct 18, 2004

descartes75

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???

2. Oct 18, 2004

Deeviant

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. Oct 19, 2004

gerben

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. Oct 19, 2004

descartes75

Thanks Guys.
But isn't 45 degree for maximum range (HOrizontal distance) ??
In this case the range is fixed (50) so intuitively the angle doesnt 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 wont ignore intuition and throw at a 45 degree angle (and hit the door).

5. Oct 19, 2004

Staff: Mentor

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 $\theta$. Find the value of $\theta$ that maximizes this function. (Take a derivative.)