Question

Tow friends traveling in Italy are doing a free-fall experiment with two balls at a non -leaning tower in Pisa. One friend is at the top of the tower and drops a ball to another friend 20.0 meters below. The friends below throw a ball straight up for her friend above to catch. With practice, the thrower can consistently achieve the right initial speed to barely reach her friend above. The students next drop and throw their balls at the same moment of time so that they pass each other in flight. What is the height is the at which the balls pass, and what are their respective speeds at this point?   

Answer 1

Suppose both balls passes each other at 'x' m above 2nd friend's position and at time 't', then

Using 2nd kinematic equation for 1st friend's drop

h1 = U1*t + (1/2)*a*t^2

U1 = initial speed of ball = 0 m/sec

h1 = height at which both balls crosses = -(20.0 - x) (since ball is 20 m above ground, So position of ball w.r.t. 1st friend will be (20 - x) m, and ball is traveling downward, So negative sign)

a = acceleration due to gravity = -9.81 m/sec^2

-(20.0 - x) = 0*t - (1/2)*9.81*t^2

Using 2nd kinematic equation for throw from ground

h2 = U2*t + (1/2)*a*t^2

U2 = initial speed of 2nd friend's throw = ?

h2 = height at which both balls crosses = x

a = acceleration due to gravity = -9.81 m/sec^2

x = U2*t - (1/2)*9.81*t^2

Now subtract both equation:

20.0 = U2*t

t = 20/U2

Now to find U2, use 3rd kinematic equation for 2nd friend's throw from ground

V2^2 = U2^2 + 2*a*d

V2 = final speed of ball when it reaches 1st friend above tower = 0 m/sec

d = total distance traveled by ball = 20 m

a = -9.81 m/sec^2

So,

0^2 = U2^2 - 2*9.81*20

U2 = sqrt (2*9.81*20) = 19.81 m/sec

So,

t = time when both balls crosses each other = 20/U2 = 20/19.81 = 1.0096 sec

Now at this time position of sandwich will be, using 1st equation

x = U2*t - (1/2)*9.81*t^2

x = 19.81*1.0096 - (1/2)*9.81*1.0096^2

x = 15.00 m = Height at which both balls crosses

Speed of 1st friend's drop ball will be: Using 1st kinematic equation:

V1 = U1 + a*t = 0 + (-9.81)*1.0096

V1 = -9.90 m/sec (-ve sign means speed is downward) (for speed use +ve values)

Speed of 2nd friend's throw ball will be: Using 1st kinematic equation:

V2 = U2 + a*t = 19.81 + (-9.81)*1.0096

V2 = +9.90 m/sec (+ve sign means speed is upwards)