# Projectiles launched at an angle

## 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.

Related Introductory Physics Homework Help News on Phys.org
G01
Homework Helper
Gold Member
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.

Ok but I dont 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
Homework Helper
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!)

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

hage567
Homework Helper
Looks good so far. What would you do next?

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 Im not sure if you can have cosX or sinX by itself?

hage567
Homework Helper
That answer looks OK to me.

It was thank you! I appreciate the help!