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) How long does the ball take to travel the first half of that distance? 03% (b) How long does the ball take to travel the second half? t2 (c) How far does the ball fall freely during the first half? h1 (d) How far does the ball fall freely during the second half? h2

Answer 1

initial vertical speed = 0

initial horizontal speed =v

the acceleration in vertical direction is a y = - g

the acceleration in horizontal direction is a x = 0

the total horizontal distance is 'd'

(A)

Finding t ₁

so in first case given horizontal direction a x = 0 so

for half of the distance travelled is

d/2 = v * t ₁

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

(B)

Finding how long does the ball taken to travel second half t ₂

The second half is also same distance with zero acceleration so

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

(c)

How long does the ball fall freely during the first half

Finding h ₁

In vertical direction we use the equation

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

Here s = - h1,  

u = 0   and

a = a y = -g

h₁ = - 1/2 * g * t 1^2

h₁ = (g * t 1^2) / 2

substituting from equation (1) we get

h1=(g×d 2)/(g v 2)

d) How far does the ball fall freely during second half,

finding h ₂

The total vertical distance is 'h'

h = 1/2 × g × (t 1 + t 2)^2

The second half vertical distance is

h ₂ = h - h 1 = [1/2 × g × (t 1 + t 2)^2] - [1/2 × (g ×t 1^2)]

h ₂ = 1/2 g [(t 1 + t 2)^2 - t 1^2]

h ₂ = 1/2 g (t 1^2 + t 2^2 + (2 t ₁ t ₂) - t 1^2)

h ₂ = 1/2 g (t ₁^2 + (2*t ₁*t ₂))

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

h 2=(3 g d 2)/(g v 2)