A baseball pitcher throws a ball horizontally at a speed of 31.6 m/s. A catcher is 18.8 m away from the pitcher. Find the magnitude of the vertical distance that the ball drops as it moves from the pitcher to the catcher. Ignore air resistance.

We apply the concepts related to the description of the movement through the kinematic trajectory equations, which include displacement, velocity and acceleration.

The trajectory equation from the motion kinematic equations is given by

[tex]y = \frac{1}{2} at^2+v_0t+y_0[/tex]

Where,

g = acceleration due to gravity

t = time

v₀ = Initial velocity

y₀ = initial position

In addition to this we know that speed, speed is the change of position in relation to time. So

[tex]v = \frac{x}{t}[/tex]

x = Displacement

t = time

[tex]t = \frac{18.8}{31.6}t = 0.5949s[/tex]

Now we know that height is given by

[tex]y =ut+\frac{1}{2}gt^2=0\times t+\frac{1}{2}\times 9.8\times 0.5949^2=1.73m[/tex]

So vertical distance will be 1.73 m

adedayo9 adedayo9 · Answer 2 · 2020-03-11T02:49:56+01:00

Answer:

s = 1.71m

Explanation:

Ignoring air resistance the horizontal velocity is a constant 31.6 m/s, since we know the distance is 18.8 m we can calculate how long it takes the ball to travel horizontally.

s = ut + 1/2at²

u = 31.6 m/s

since the ball is travelling horizontally a =0, therefore

s = ut

t = s/ u

t = 18.8 / 31.6 ==0.59s

Now consider vertical motion

s = ut + 1/2gt²

Since the acceleration is going down ( acceleration due to gravity is positive +g) and the initial y velocity is zero.

so we have that s = [tex]\frac{1}{2}[/tex]gt²

s = 1/2 × 9.8 × 0.59²

=[tex]\frac{1}{2}[/tex] ×9.8×0.3481

s = 1.71m