Projectile motion (Baseball Field)

Question

A baseball is thrown from the outfield to home plate. Let's say the outfielder and catcher are 71.6m horizontally apart, and the ball leaves the outfielders hand at a height of 2.28m. (Assume no air resistance)

 If it takes 1.94s for the ball to get from fielder to catcher, what was the magnitude of the velocity of the ball in the "x" right before the catcher gets the ball?

If the catcher catches the ball at a height of 2.17m, find the vertical velocity the ball had when it left the fielders hand.

At what angle (in degrees) did the fielder throw the ball with respect to the ground?

Answer 1

Let

d = 71.6 m be the distance between the outfielder and the catcher,

h₁ = 2.28 m be the height at which the ball is thrown,

h₂ = 2.17 m be the height at which the ball is caught,

h (unknown) be the maximum height the ball reaches,

t = 1.94 s be the time the ball is flying,

t₁ (unknown) be the time for the ball to reach the maximum height after the throw,

t₂ (unknown) be the time for the ball to fly from the maximum height to the catcher,

g = 9.81 m/s² be gravitational acceleration,

v_x (to be computed) be the horizontal velocity,

v_y (to be computed) be the vertical velocity at the throw,

θ (to be computed) be the angle of throw.

Horizontal velocity:

Horizontal velocity is constant because there is no horizontal force to cause any horizontal acceleration.

Therefore

v_x = d/t = 71.6/1.94 = 36.9 m/s

Vertical velocity:

For time t₁ the ball travels upward total vertical distance h - h₁, and then

for time t₂ it travels downward total vertical distance h - h₂.

We have the following kinematic equations:

h - h₁ = gt₁²/2 (1)

h - h₂ = gt₂²/2 (2)

v_y = gt₁ (3)

In addition

t₁ + t₂ = t (4)

So we have four equations and four unknowns -- h, t₁, t₂, v_y.

Combine (1) and (2) eliminating h:

gt₁²/2 + h₁ = gt₂²/2 + h₂

Manipulate to get t₁ and t₂ on the left side

t₁² - t₂² = 2(h₂ - h₁)/g

Rewrite t₁² - t₂²

(t₁ + t₂)(t₁ - t₂) = 2(h₂ - h1)/g

Substitute from (4)

t(t₁ - t₂) = 2(h₂ - h₁)/g

Manipulate to get t₁ and t₂ on the left side

t₁ - t₂ = 2(h₂ - h₁)/gt (5)

Add (4) and (5)

2t₁ = t + 2(h₂ - h₁)/gt

Express t₁

t₁ = t/2 + (h₂ - h₁)/gt

Substitute into (3)

v_y = gt/2 + (h₂ - h₁)/t

Substitute actual numbers

v_y = 9.81×1.94/2 + (2.17 - 2.28)/1.94 = 9.46 m/s

Angle of throw:

tan(θ) = v_y/v_x

θ = arctan(v_y/v_x) = arctan(9.46/36.9) = 14.3°