Question

A particle is thrown with a velocity of u m/s . It passes a pt A and B having same height at time t1 = 1 sec and t2 = 3 sec . what is the value of u ?

Answer 1

value of u = 19.6 m/s

explanation : initial speed of particle = u m/s

particles passes points A and B having same height at time t1 = 1sec and t2 = 3sec .

using formula, s = ut + 1/2 at²

here, a = -g , t = 1sec

then, s = u(1) + 1/2 (-g)(1)²

s = u - 1/2g .........(1)

again, putting a = -g , t = 3sec

s = u(3) + 1/2 (-g)(3)²

s = 3u - 9g/2 .........(2)

from equations (1) and (2),

u - 1/2g = 3u - 9g/2

⇒9g/2 - 1/2 g = 3u - u = 2u

⇒(9g - g)/4 = u

⇒u = 2g

as we know, g = 9.8 m/s²

so, u = 2 × 9.8 m/s = 19.6 m/s

Answer 2

$\huge\bold\purple{Answer:-}$

initial speed of particle = u m/s

particles passes points A and B having same height at time t1 = 1sec and t2 = 3sec .

using formula, s = ut + 1/2 at²

here, a = -g , t = 1sec

then, s = u(1) + 1/2 (-g)(1)²

s = u - 1/2g .........(1)

again, putting a = -g , t = 3sec

s = u(3) + 1/2 (-g)(3)²

s = 3u - 9g/2 .........(2)

from equations (1) and (2),

u - 1/2g = 3u - 9g/2

⇒9g/2 - 1/2 g = 3u - u = 2u

⇒(9g - g)/4 = u

⇒u = 2g

as we know, g = 9.8 m/s²

so, u = 2 × 9.8 m/s = 19.6 m/s