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*t^{2}+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)t^{2}+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 t^{2}=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.

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