Question

a stone is thrown vertically upward with a speed of 40 metre per second the time interval for which particle of about 40 m from the ground is dash​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Time\:taken=6.8\:sec}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a stone is thrown vertically upward with a speed of 40 m/s.

• We have to find the time interval at a height of 40 m from the ground.

$\underline \bold{Given : } \\ \implies Initial \: velocity(u) = 40 \: m/s \\ \\ \implies Distance (s)= 40 \: m \\ \\ \implies Acceleration(a) = - 10 \: m /{s}^{2} \\ \\ \underline \bold{To \: Find : } \\ \implies Time \:taken(t)= ?$

• According to given question :

$\bold{Using \: second \: equation \: of \: motion :} \\ \implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {v}^{2} = {40}^{2} + 2 \times ( - 10) \times 40 \\ \\ \implies {v}^{2} =1600 - 800 \\ \\ \implies {v}^{2} = 800 \\ \\ \implies v = \sqrt{800} \\ \\ \bold{\implies v = 20 \sqrt{2} \: m/s}\\ \\ \bold{Using \: first \: equation \: of \: motion : } \\ \implies v = u + at \\ \\ \implies 20 \sqrt{2} = 40 + ( - 10) \times t \\ \\ \implies 20 \sqrt{2} - 40 = - 10t \\ \\ \implies t = \frac{20 \sqrt{2} -40 }{ - 10} \\ \\ \implies t = \frac{ - 10(2 \sqrt{2} + 4) }{ - 10} \\ \\ \implies t = 2 \times 1.4 + 4 \\ \\ \bold{\implies t = 6.8 \: sec}$

Answer 2

$\implies {v}^{2} = {u}^{2} + 2as \\ \\ \implies {v}^{2} = {40}^{2} + 2 \times ( - 10) \times 40 \\ \\ \implies {v}^{2} =1600 - 800 \\ \\ \implies {v}^{2} = 800 \\ \\ \implies v = \sqrt{800} \\ \\ \bold{\implies v = 20 \sqrt{2} \: ms}\\ \\ \implies v = u + at \\ \\ \implies 20 \sqrt{2} = 40 + ( - 10) \times t \\ \\ \implies 20 \sqrt{2} - 40 = - 10t \\ \\ \implies t = \frac{20 \sqrt{2} -40 }{ - 10} \\ \\ \implies t = \frac{ - 10(2 \sqrt{2} + 4) }{ - 10} \\ \\ \implies t = 2 \times 1.4 + 4 \\ \\ \bold{\implies t = 6.8 \: sec}$