Projectile motion - velocity/time unknown

[zombiemegan22]

A batter hits a pitched ball at a height of 4 feet above the ground so that its angle of projection is 45 degrees and its horizontal range is 350 feet.
The ball is fair down the left field line where a 24 foot high fence is located 320 feet from home plate.
By how much will the ball clear the fence?

vfy=viy+at
(vf)$$^{2}$$= (vi)$$^{2}$$+2aD (..we'll say D= displacement)
D= vi(t)+.5(a)(t$$^{2}$$)

Vx=Dx(t)

Well, I only really know my x displacement, my acceleration constant, and that my initial y velocity and my x velocity are the same because of the 45 degree triangle [at least that's what my teacher said]...

So, help?! It's been bugging me for days.

 

[physics girl phd]

Once you have your components of initial velocity along X and Y, X (horizontal) and Y (vertical) equations of motion are only related by time since there is no force along X and the gravitational force along Y, and no constraints on the object like a plane or a string, etc. (it's in free-fall) ... therefore:

First find the time when the ball is passing the fence.

Using this time, find the Y location (the vertical displacement).

Since the fence is only so tall, you can know if it clears and by how much.

P.S. Welcome to the forums!

 

[thomate1]

Based on the given information, we can use the equations of projectile motion to calculate the velocity and time of the ball. First, we can use the horizontal range (350 feet) to find the time of flight (t) using the equation Vx=Dx/t. Since the initial and final velocities in the horizontal direction are the same, we can simplify this equation to Vx=Dx/t. Plugging in the given values, we get t=350 feet/Vx.

Next, we can use the height of the fence (24 feet) to find the vertical displacement (Dy) using the equation Dy=viy(t)+1/2(-9.8 m/s^2)(t^2). Since the initial vertical velocity (viy) is equal to the final velocity (0 m/s) at the highest point of the ball's trajectory, we can simplify this equation to Dy=-4.9(t^2). Plugging in the value for t that we found in the previous step, we get Dy=-4.9(350 feet/Vx)^2.

Now, we can use the given angle of projection (45 degrees) to find the initial velocity (vi) using the equation vi=Vx/cos(45). Plugging in the value for Vx that we found in the first step, we get vi=(350 feet/t)/cos(45).

Finally, we can use the equation Vfy=viy+at to find the final vertical velocity (Vfy) at the point where the ball reaches the top of the fence. Since we know that the initial vertical velocity (viy) is 0 m/s, we can simplify this equation to Vfy=-9.8(t). Plugging in the value for t that we found in the first step, we get Vfy=-9.8(350 feet/Vx).

Now, to answer the question of how much the ball will clear the fence, we need to find the difference between the height of the ball at the top of the fence (Dy) and the height of the fence (24 feet). So, the ball will clear the fence by a distance of Dy-24 feet, which is equal to -4.9(350 feet/Vx)^2-24 feet.

In conclusion, using the equations of projectile motion, we can calculate that the ball will clear the fence by a distance of -4.9(350 feet