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 (vo) of ball A = 10 m/s
Time interval (t) of ball A = x
Initial speed (vo) of ball B = 25 m/s
Time interval (t) of ball B = x – 1
Acceleration due to gravity (g) = -10 m/s2 (given minus sign because the direction of gravity is downward)
Wanted : The height of ball B when it encounters ball A (h)
[irp]
hA = hB
vo t + ½ g t2 = vo t + ½ g t2
10x + ½ (-10) x2 = 25 (x-1) + ½ (-10) (x-1)2
10x – 5x2 = 25 (x-1) – 5 (x-1)2
10x – 5x2 = 25x – 25 – 5 (x2-2x+1)
10x – 5x2 = 25x – 25 – 5x2 + 10x – 5
10x – 5x2 – 25x + 25 + 5x2 – 10x + 5 = 0
– 5x2 + 5x2 + 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 = vo t + ½ g t2 = (10)(1.2) + 1/2 (-10)(1.2)2 = 12 – 5(1.44) = 12 – 7.2 = 4.8 meters
The height of ball B when it encounters ball A :
h = vo t + ½ g t2 = (25)(0.2) + 1/2 (-10)(0.2)2 = 5 – 5(0.04) = 5 – 0.2 = 4.8 meters
[irp]
