Minimum speed for baseball to clear 12m fence 95m away

[pyromaniac2]

Hey, have stared at this problem for a good half hour now, and my brain must be completely shot by now so I am having difficulty solving. If anyone can point me in the right direction, I would greatly appreciate it. Thanks!

When Babe Ruth hit a homer over the 12m-high right field fence 95m from home plate, roughly what was the minimum speed of the ball when it left the bat? Assume the ball was hit 1.0m above ground and its path initially made a 40 degree angle with the ground.

 

[Fermat]

Let V be the initial velocity of the ball.
You know the angle it was hit at so you can resolve this velocity into horizontal and vertical components.
Since you now know the horizontal velocity, then what is the time taken, in terms of V, to travel the 95m to the fence ?
It will take the same amount of time for the ball to travel vertically from a height of 1 m up to a maximum height then descend until it is at a height of 12m above the ground.

Do you understand what I'm saying in the last sentence ?

This should allow you to solve your problem.

 

[pyromaniac2]

Attempt #1

Well, I think so. Although let me show you kind of what I tried to do with this and see how far off it is. I broke it into Vx and Vy with Vx= 95cos40 and Vy= 11sin40. I then took the 95 meters and divided it by the Vx (roughly 72.77) to get a time of (roughly) 1.305 seconds. I then put it all into the formula:

Change in position = V(t) + 1/2a(t)^2

To get a result somewhere along the lines of 78.16m/s

But that's just kind of a stab in the dark based on what I thought you were advising. Close or am I not even in the ballpark? (no pun intended)

 

[Fermat]

In the outfield, I would say!

Vx = V.cos40
Vy = V.sin40

Vx = 95/t

where t is the time to travel 95 m horizontally.

You should now have t in terms of V.

It looks like you were doing the vertical displacement bit right with the correct formula. You should substitute for t into that formula and then solve for V.