Baseball Problem: Does a Ball Hit the Roof? Answer Here!

[jfeelz33]

Homework Statement

This is part of a calculus test and I am clearly missing something because this should be an easy problem: A baseball pitcher is having fun with the fans. He can throw a ball vertically froma point 6 feet above the ground at 75 miles per hour (110 ft/sec). He is standing in a domed stadium with a roof 200 feet above the ground. Does the ball hit the roof? Justify your answer.

We are currently doing integrals and just finished derivatives if that's any help. Any help would be appreciated, thanks.

The Attempt at a Solution

I sadly don't have one.

 

[Doc Al]

Clearly you need to figure out the maximum height the ball reaches. Can you write an expression for the height as a function of time?

You can start with the acceleration due to gravity and integrate if you like.

 

[ogb p]

I would approach this problem by first defining the variables involved and setting up the relevant equations. The key variables here are the height of the ball, the height of the roof, and the initial velocity of the ball. From these, we can use the equations of motion to determine the position of the ball at any given time.

Using the equation h(t) = h0 + v0t - (1/2)gt^2, where h(t) is the height of the ball at time t, h0 is the initial height, v0 is the initial velocity, and g is the acceleration due to gravity, we can set up an equation for the height of the ball at any given time.

In this problem, we are given that the initial height (h0) is 6 feet, the initial velocity (v0) is 110 ft/sec, and the height of the roof is 200 feet. We can also assume that the ball is thrown straight up, so the acceleration due to gravity (g) is negative and equal to -32 ft/sec^2.

Now, to determine if the ball hits the roof, we need to find the time at which the height of the ball is equal to the height of the roof. We can set up the following equation:

200 = 6 + 110t - (1/2)(32)t^2

Solving for t, we get t = 4.08 seconds. This means that after 4.08 seconds, the height of the ball will be equal to the height of the roof. Therefore, the ball does hit the roof.

In terms of calculus, we can also use the concept of limits to prove this. The height of the ball at any given time can be represented by a function, h(t), which we can take the derivative of to find the velocity of the ball at any given time. We can then take the limit of this velocity function as t approaches 4.08 seconds to determine the exact moment when the ball hits the roof.

In conclusion, using both equations of motion and calculus, we can determine that the ball does hit the roof in this scenario. It is important to always justify our answers with relevant equations and logical reasoning in scientific problems.