How far will a baseball travel if it is thrown or hit 80 mph at an 45 degree angle?

Question

We have a radar gun at home and we can measure the velocity of the baseball when it leaves a pitchers hand, or when it leaves the bat after contact. We can also roughly estimate the angle or trajectory of the ball. I am sure there are equations out there that involve takeoff velocity in conjunction with the angle of takeoff, and also air resistance equations. I'm hoping someone with the knowledge or info can help me out. Basically , want to know how far the ball will travel when I hit it 80miles per hour (ball of the bat speed) and I hit it at a 45 degree angle. Thanks...

A little more info...thanks so far for the great input... A baseball weighs 5 ounces, it is a sphere, and it has a circumference of 9 inches.. Can anyone calculate air resistance with that data? Lets assume that the ball is hit in a dome ( no wind, perfectly calm) and it is at sea level... This question seems to get more complicated! The original answers of 428 feet seems almost unbelievable to me with an inital ball velocity of 80mph. My personal maximum throwing velocity is 80mph, and I know for a fact that I cannot throw a baseball more than 350 feet on a calm day with an optimum release angle of about 45 degrees. Oh yeah, estimating a batting ball distance, the ball is usually about 3 feet off the ground when hit (and about 5.5 feet off ground when thrown). I actually simply guessed at the answer to this question and came up with about 330-350 feet, so the 428 foot answer is blowing my mind right now. Thanks again for any further input.

Answer 1

1. You need to know the initial height of the ball from the ground. I will assume from ground level.

2. Calculate the vertical and horizonal components of the speed.
a^2 + b^2 = c^2 where c = 80 and since it is a 45 angle a=b
2a^2 = 80^2
a = 56.6 mph (both vertical and horizontal speeds are the same)

2. Now take the vertical speed (56.6mph) and find time it takes to hit the ground.
Since acceleration due to gravity is -32.2 ft/(s^2)
56.6 mi/h * 1/60 h/min * 1/60 min/s * 5280 ft/mi = 83 ft/s
Now using
a=(v2-v1)/t where a is acceleration, v1 = inital velocity, and v2 = final velocity, and t = time it took
-32.2 = (0 - 83)/t
t = -83 / -32.2 = 2.58sec
since that is the time for the ball to finally reach the top of the arc, multiply by two to get the complete arc
total time in the air = 5.16sec

3. Calculate distance travelled in that time.
I am leaving air resistance out due to incomplete data (you need mass of the ball, resistance will also change due to the shape, the barometric pressure, etc.)
5.16s * 83 ft/s = 428.3 ft travelled

Edit:
Since the distance from the ground is small compared to the velocity of it falling, we can estimate just an additiona 0.04 secs so total time is now 5.20 secs

4. Drag equation
D = Cd * A * 0.5 * r * V^2
Cd = Drag coefficient
A = reference area (half surface area of baseball 2pi * r^2 = 3.53 ft^2)
r = density of air (20 C and sea level is 1.2 kg/m3)
r = 1.2 kg/m^3 * 2.2 lb/kg * 0.0283 m^3/ft^3 = 0.0747 lb/ft^3

Cd can't be calculated since it is different for every object and situation. Only way for Scientist to find it is using a wind tunnel.
I cheated and looked it up for a standard baseball it is 0.007 lb/ft
D = 0.007 * 3.53 * 0.5 * 0.0747 * V^2
D = 0.0009229 * V^2
Di = 0.0009229 * 83^2 initial force from drag
Di = 6.36
use F = ma to get the negative initial acceleration
6.36 = 0.3125 * ai
ai = 20.4

-20.4 = (d/5.20 - 83)/5.20
d = 120 ft
This is only true if the drag stays constant, but it decreases as the velocity decrease (exponentially).

An estimation can be done by calcing the acceleration after each second, then decreasing velocity due to that acceleration.
V(i) = 83.0, Drag(i) = 6.36, a(i) = -20.35
V(1) = 62.7, Drag(i) = 3.62, a(i) = -11.59
V(2) = 51.1, Drag(i) = 3.41, a(i) = -7.70
V(3) = 43.4, Drag(i) = 1.74, a(i) = -5.55
V(4) = 37.8, Drag(i) = 1.32, a(i) = -4.22
V(f) = 33.6
After adding up the distances travelled after each second (including the 0.2 sec at the final velocity), I come up with an estimate of 284.6 ft travelled.

This is still a desent estimate, but there is still additional factors to consider. The ball may reach terminal velocity on its way down which will increase the time returning to the ground. This problem at first glance looks simple, but as you can see, it gets complex very quickly.

Answer 2

Let's use metric units, which is easier. 80 mph = 35.76 m/s. The upward component of that velocity is sin(45) * 35.76 = 25.29 m/s.

First question: how long before the ball lands? Since the accelation due to gravity is 9.8 m/sec^2, we compute 25.29 / 9.8 = 2.58 seconds until the ball's vertical velocity is zero (at the top of its arc), and another 2.58 seconds until it comes down, for a total of 5.16 seconds.

Second question: how far will the ball travel HORIZONTALLY in 5.16 seconds? To make it easy, we will ignore air resistance. The horizontal component of the ball's velocity is cos(45) * 35.76, which is also 25.29 m/s (because it's at a 45 degree angle). So the distance travelled will be 25.29 * 5.16, or 130.5 meters. That's 428 feet.

Answer 3

I think the equation comes from the impact.
The force of the bat upon impact (Newton)= distance travelled.( ma)+ sound energy ( don't know) + air friction ( don't know) + wind resistance ( different height, different velocity) + skin friction of the ball ( like the world cup ball, more aero dynamic)...maybe...

Answer 4

Wow! this is tough.. first of all.. the threads on a baseball cause it to curve as it spins through the air.. so any motion has to take into account how fast the baseball is spinning and what direction.

Answer 5

I dont remember the exact formula, but the ball would be so affected by wind and humidity that you cant really get an accurate answer.