A pitcher throws a baseball from a height of 2.5 meters with a velocity of 21 m/s. How long before the ball hits the ground? How far will the ball travel?

Question

Free Fall Kinematic Equations which may help with this problem.

Vf=Vi + 9.8*t

 

d=Vi*t + .5*9.8*t²

 

Vf² = Vi² + 2*9.8*d

 

d= [Vi + Vf]*t/2

Answer 1

Assuming that the baseball is thrown perfectly horizontally.

First, find the time it takes for the ball to hit the ground. As the ball is thrown horizontally, the initial velocity in the vertical direction is 0, i.e. v_y,0 = 0 m/s. Using the acceleration of gravity, a = -9.81 m/s², we can use the kinematic equation d = v_y,0*t + at²/2. Using d_y = -2.5 m (assuming that the origin is the hand of the pitcher, the ball goes below that point 2.5 m), a = g = -9.81 m/s², we solve for t:

1. -2.5 = 0*t + -9.81*t²/2 = -4.905*t²
2. t = 0.714 s

Now, we know that the velocity in the x direction is 21 m/s, i.e. v_x = 21 m/s. Assuming that the velocity stays constant, then we can find the distance traveled in the x direction as the velocity multiplied by the time, i.e.

1. d_x = v_x*t = 21*0.714 = 15 m

Answers:

1. t = 0.714 s
2. d_x = 15 m

Answer 2

Hadassah,

This is one of three consecutive problems posted that utilize the same process to solve. I presume that if I explain this one correctly, you'll be able to solve the other two.

First, a couple of assumptions that are unstated, but probably implied:

1) The ball is released from a point 2.5 meters above the ground. I.e., the height of the release is measured from the ground, not from a pitching mound.

2) The ball is released horizontally, not at an angle above or below the horizontal.

3) Air resistance on the ball is negligible -- not considered.

Okay, first question: How long before the ball hits the ground?

Once the ball is released, the only force acting on it is gravity. Therefore the applicable equation is:

d=Vi*t + .5*9.8*t^2 Where the initial vertical velocity is zero. Plug in 2.5 for d, and solve for t. I get

t-= 0.71 seconds. This is the time it takes to drop to the ground while also travelling forward,

Second question: How far will it travel?

Since we are neglecting air resistance, the horizontal velocity is constant at 21 m/s. The ball will travel at that speed until gravity brings it to ground after 0.71sec

So, d = v*t = 21 m/s * 0.71 sec = 14.91 meters.

You can use this exact process to solve the other two problems. HINT: The bullet should go farther than the ball......