Projectiles launched at an angle

[aly1201]

Homework Statement

A baseball is hit over a 7.5-m-high fence 95 m from the home plate. What was the minimum speed of the ball when it left the bat? Assume the ball was hit 1 m above the ground and its path initially made a 38 degree angle with the ground.

Homework Equations

y=v(initial, y direction)t-1/2at^2 x=v(initial, x direction)t

The Attempt at a Solution

t=95/v 95/v(cos38) ? plug 95/v(cos38) into the y equation? and substitute v(sin38) for initial y velocity? This is where I get lost; our teacher wrote all the work on the board very quickly and came up with an answer somehow, but when I did the work I didnt even come close.

 

[G01]

You have the following (correct) equation for the height of the ball right when it gets to the fence:

y=v\sin(38^o)-\frac{1}{2}g\left(\frac{95}{v\cos(38^o)}\right)^2

Now how high do you want the ball to be so that it just barely the fence? Plug that in for y and when you solve for v, you'll get the minimum speed.

 

[aly1201]

Ok but I don't understand how to get v by itself, with the cos38 and sin38 there. And in the equation you wrote is "t" supposed to be in there at all?

 

[hage567]

There is a t missing in the vsin(38) term in that equation. G01 has substituted for t since you can express it in terms of the horizontal distance and horizontal velocity. As already pointed out, you can figure out what y should be from the information given in the problem. So the equation should look like this:

y=v\sin(38^o)\left(\frac{95}{vcos(38^o)}\right)-\frac{1}{2}g\left(\frac{95}{v\cos(38^o)}\right)^2

Now it is just algebra. To isolate v, start by getting the term containing v on its own on one side of the equation (note the v's cancel in the first term!)

 

[aly1201]

Ok so [95/v(cos38)]^2=[6.5-(tan38)(95)]/-4.9? because sinX/cosX=tanX

 

[hage567]

Looks good so far. What would you do next?

 

[aly1201]

I solved the right side and got 13.8, which i took the square root of (since the left side was squared) and got 3.7. Then I divided that by 95 and got 25.6. So now v(cos38)=25.6? If it does then I could just say 25.6/cos38=v, so 25.6/.79=32.4?
I might've rounded too much so it might not be exact, and I am not sure if you can have cosX or sinX by itself?

 

[hage567]

That answer looks OK to me.

 

[aly1201]

It was thank you! I appreciate the help!