A baseball leaves a pitcher\'s hand horizontally at a speed v. The distance to t

Question

A baseball leaves a pitcher's hand horizontally at a speed v. The distance to the batter is d. (Ignore the effect of air resistance. Use any variable stated above along with the following as necessary: g for acceleration due to gravity.)

a. t1 = How long does the ball take to travel the first half of that distance?

b. t2 = How long does the ball take to travel the second half?

c. h1 = How far does the ball fall freely during the first half?

d. h2 = How far does the ball fall freely during the second half?

Explanation / Answer

initial horizontal speed is 'v'

initial vertical speed is zero

the acceleration in horizontal direction is ax = 0

the acceleration in vertical dirction is ay = - g

the total horizontal distance is 'd'

a)

in horizontal direction ax = 0 so

for half of the distance

d/2 = v * t1

t1 = d / (2v)----------------(1)

b)

the second half is also same distance with zero acceleration so

t2 = d / (2v)----------------(2)

c)

In vertical direction we use the equation

s = ut + (1/2 * a * t^2)

here s = - h1,    u = 0   and a = ay = -g

- h1 = - 1/2 * g * t1^2

h1 = (g * t1^2) / 2

substituting from equation (1) we get

h1 = (g * d^2) / (8*v^2)

d)

the total vertical distance is 'h'

h = 1/2 * g * (t1 + t2)^2

the second half vertical distance is

h2 = h - h1 = [1/2 * g * (t1 + t2)^2] - [1/2 * (g * t1^2)]

h2 = 1/2 * g * [(t1 + t2)^2 - t1^2]

h2 = 1/2 * g * (t1^2 + t2^2 + (2*t1*t2) - t1^2)

h2 = 1/2 * g * (t2^2 + (2*t1*t2))

substituting from equation (1) and (2) we get

h2 = (3 * g * d^2) / (8 * v^2)

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