1. Ball A threw vertically upward with the speed of 10 m/s. 1 second later, from the same position, Ball B is thrown vertically upward at the same path, with the speed of 25 m/s. What is the height of ball B when it encounters ball A.

Solution :

In solving the problem of vertical motion, the vector quantity which direction upward is given a positive sign, the vector quantity that direction downward is given a negative sign.

__Known :__

Initial speed (v_{o}) of ball A = 10 m/s

Time interval (t) of ball A = x

Initial speed (v_{o}) of ball B = 25 m/s

Time interval (t) of ball B = x – 1

Acceleration due to gravity (g) = -10 m/s^{2 }(given minus sign because the direction of gravity is downward)

__Wanted :__ The height of ball B when it encounters ball A (h)

[irp]

h_{A} = h_{B}

v_{o} t + ½ g t^{2 }= v_{o} t + ½ g t^{2}

10x + ½ (-10) x^{2 }= 25 (x-1) + ½ (-10) (x-1)^{2}

10x – 5x^{2 }= 25 (x-1) – 5 (x-1)^{2}

10x – 5x^{2 }= 25x – 25 – 5 (x^{2}-2x+1)

10x – 5x^{2 }= 25x – 25 – 5x^{2 }+ 10x – 5

10x – **5x**^{2} – 25x + 25 + **5x**^{2} – 10x + 5 = 0

– 5x^{2} + 5x^{2 }+ 10x – 25x – 10x + 25 + 5 = 0

10x – 25x – 10x + 25 + 5 = 0

– 25x + 25 + 5 = 0

– 25x + 30 = 0

– 25x = – 30

x = -30/-25

x = 1.2 seconds

Time interval ball A in air before it encounters ball B = 1.2 seconds

Time interval ball B in air before it encounters ball A = 1.2 seconds – 1 seconds = 0.2 seconds.

The height of ball A when it encounters ball B :

h = v_{o} t + ½ g t^{2 }= (10)(1.2) + 1/2 (-10)(1.2)^{2 }= 12 – 5(1.44) = 12 – 7.2 = **4.8 meter****s**

The height of ball B when it encounters ball A :

h = v_{o} t + ½ g t^{2 }= (25)(0.2) + 1/2 (-10)(0.2)^{2 }= 5 – 5(0.04) = 5 – 0.2 = **4.8 meter****s**

[irp]