Am I missing something in this problem...?

Question

I'm working on a physics problem, which seems to be a "changing y" problem. However, it feels like something is missing from the question...can someone point it out to me? Here's the original question:

"When Babe Ruth hit a homer over the 12 m high right field fence 100m from home plate, roughly what was the minimum speed of the ball when it left the bat? Asume the ball was hit 1.0 m above the ground and its path initially made a 40 degree angle with the ground."

IF this was a changing Y problem, I would assume: Dy = 11m, Dx = 100m, Initial Vx = Initial V[(cos400], Initial Vy = Initial V[(sin400], a = -9.8 m/s^2.

Is there something I'm just missing? Please help!

Answer 1

It is a projectile type question. You have the initial angle of the projectile and the range and height (coordinates of a point on the parabola or flight trajectory.) The minimum is because you can draw a trajectory that will just clear the fence

Answer 2

Here is the method:

The ball leaves the bat with initial velocity v0. This velocity is made up of two components, v0x (horizonta) and v0y (vertical). The horizontal compnent does not change, but the vertical component is affected by gravity. Therefore the trajectory is described by vx(t) =v0x, and vy(t) = v0y - gt, where = accel due to gravity (9.8 m/sec^2). The vertical height y(t) of the ball is the integral [vy(t)*dt]. This is v0y*t - .5*g*t^2 + C. The ball starts at an elevation of 1.0m, so y(0) = 1, C = 1. Therefore the formula for vertical height is y(t) = v0y*t - ,5*g*t^2 + 1.

To reach the horizontal distance of 100m, the ball must travel for tr = 100/v0x. The vertical height after travelling tr you get from y(tr), y(100/v0x) = 100*(v0y/v0x) - .5*g*(100/v0x)^2 + 1, and this must exceed 11m.

The ball leaves the bat at an angle of 40 deg, so

v0y/v0x =tan (40deg); substituting this in for v0y/v0x above

100*tan(40) - ,5*g*(100/v0x)^2 + 1 = 11

100*tan(40) - .5*g*(100/v0x)^2 = 10

.5*g*(100/v0x)^2 = 100*tan(40) - 10
(100/v0x)^2 = (100*tan(40) - 100)/(.5*g)
100/v0x = sqrt[(100*tan(40) - 100)/(.5*g)]
1/v0x = (1/100)* sqrt[(100*tan(40) - 100)/(.5*g)]
v0x = 100/ sqrt[(100*tan(40) - 100)/(.5*g)]

This is just a number--you know all the values here, so find v0x.
Then, since v0y/v0x = tan(40), you can get v0y = v0x*tan(40)

Once you have v0x and v0y, the ball velocity is the vector sum of these, and since they are at right angles, v = sqrt(v0x^2 +v0y^2).

Answer 3

No, its correct, and I will solve it from first principles

let U=initial vel.
Uy = U.sin 40 = 0.643U
Ux = U.cos40 = 0.766U

Let
t = time taken to clear the fence, 100m away, 11m above the point of projection
then
100 = Ux.t
Ux=100/t= 130.55/t.... (1)

For vertical displacement
height = Uy.t - 0.5g.t^2 = 11 .... (2)

substituting for Ux, Uy appropriately,

t = sqrt (2x(0.643x130.55-11)/9.8) = 3.858 sec,

and

U = 130.55/3.858 = 33.84 m/sec

So,
Babe Ruth was right, huh?

Answer 4

The vertical component of the balls' velocity after impact was vsin 40, and that was enough for the ball to climb 11 m against gravity, so (v sin 40)^2 = 2x9.8x11=215.6 so v sin 40= 14.68 so v=25 metres/sec.

Answer 5

Looks like you are on the right track. You are going to need:

Vx = Vo*cos(40)
Vy = Vo*sin(40)

y = Vy*t + 1/2*a*t^2
x = Vx*t

when you start making substitutions you'll also want to recognize:

tan(40) = Vy/Vx

I got Vo = 33.84 m/s (If i did my arithmetic correctly )

Answer 6

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Answer 7

Here use this equation: "velocity" squared divided by "gravity" times "2 sin theta" equals X. Manipulate the equation to answer it

Answer 8

looks like it

Answer 9

yikes. good luck