Question

two block both weighing 1 kg are shown in the figure block P started with 10 m/s and kept 10 second, whereas block Q started with 0m/s and reached 5m/s in 5seconds. indentify the force applied in each case​

Answer 1

u = 9 m/ Sec

a = - g = - 9.81 m/ sec2

t = time taken by the ball to reach the height “h”

SA = Height reached by the ball A

Equation of motion

+ ݐݑ = ݏ

ଵ

ଶ

ଶݐܽ

SA = 9 t + ½ (- 9.81) x t2

SA = 9 t - 4.905 t

2

…………(1)

Boy B : Consider the upward motion of the ball

u = 12 m/ sec

a = - g = - 9.81 m/ sec2

SB = Height reached by the ball B

SB = SA +2.5

t = time taken ( time remains same for both the cases)

Equation of motion

S = ut + ½ at2

SB = 12 t – ½ x 9.81 t2

= 12t – 4.905 t2 …………. (2)

Substitute (1) in (2)

SB = SA + 2.5 = 9 t - 4.905 t2

+ 2.5 = 12 t – 4.905 t

2

12t – 9 t = 2.5

t = 0.833 sec.

There fore

SA = 9 x 0.833 – 4.905 x 0.8332

= 4.093 m

SB = 4.093 + 2.5 = 6.593 m

Height of the two ball from ground = 6.593 m