Throw and Catch

Question:

A boy(A) throws a ball from a building horizontally with a velocity of 10 feet per second from a height of 20 feet. A second boy(B) is running straight away from the building at a speed of 4 feet per second at the instant Boy(A) throws the ball. What is the minimum horizontal distance(d) from the building of the boy(B) if he is to catch the ball thrown by boy(A) before it hits the ground? What is the relative velocity of the ball and boy(B) at the time of the catch?

Answer:

We know the expression for the position of an object subject to constant acceleration(a), as a function of time is x=1/2*a*t²+v0*t+x0. If not, see Two Ball Meeting.

The ball has 10 f/s horizontal velocity. Its vertical displacement is due to gravity so the height of the ball, h=1/2(-32)t²+20. The ball has to fall 20 feet to be caught just prior to hitting the ground allowing the minimum head start for boy(B), so it will be in the air t seconds where t²=20/16 or t=1.12 seconds. For boy(B) to catch the ball he must reach the point of impact 1.12 seconds after the ball is thrown. At 4 f/s this represents a distance of 4.48 feet. The ball will travel 10(1.12) feet horizontally while it is in the air so boy A must be 11.2-4.48 = 6.72 feet from the building when the ball is thrown. The relative velocity has a horizontal component of 10-4=6f/s. The vertical component of the relative velocity is whatever speed the ball packed up in falling for 1.12 seconds. That is 1.12*32=35.84f/s. The speed is the square root of the sum of the squares of the velocity components or 36.34f/s.

This information is brought to you by M. Casco Associates, a company dedicated to helping humankind reach the stars through understanding how the universe works. My name is James D. Jones. If I can be of more help, please let me know. JDJ