A 200gm ball thrown from a height of 200m from the surface. Find the t

1.	A 200gm ball thrown from a height of 200m from the surface. Find the time taken by the ball to touch the surface. Also, find the velocity of the ball at that time.
Answer» GIVEN : Mass of ball (m) = 200 g = 0.2 kg Height (h) = 200 m Initial velocity (u) = 0 m/s Acceleration due to GRAVITY (g) = 10 m/s² To FIND : Time taken by the ball to touch the surface. Final velocity of the ball. SolutioN : Use 2nd equation of motion : $\implies \rm{s = ut + \dfrac{1}{2} gt^2} \\ \\ \\ \implies \rm{200 = 0 \times t + \bigg( \dfrac{1}{2} \times 10 \times t^2 \bigg) } \\ \\ \\ \implies \rm{200 = 0 + 5 t^2} \\ \\ \\ \implies \rm{5t^2 = 200} \\ \\ \\ \implies \rm{t^2 = \dfrac{200}{5}} \\ \\ \\ \implies \rm{t^2 = 40} \\ \\ \\ \implies \rm{t = \sqrt{40}} \\ \\ \\ \implies \rm{t = 6.324} \\ \\ \\ \large \implies {\boxed{\rm{t \: \approx \: 6.3 \: s}}}$ ________________________ Now, use 1st equation of motion : $\implies \rm{v = u + gt} \\ \\ \\ \implies \rm{v = 0 + 10 \times 6.3} \\ \\ \\ \large \implies {\boxed{\rm{v \: = \: 63 \: ms^{-1}}}}$

A 200gm ball thrown from a height of 200m from the surface. Find the time taken by the ball to touch the surface. Also, find the velocity of the ball at that time.

Answer»

GIVEN :

Mass of ball (m) = 200 g = 0.2 kg
Height (h) = 200 m
Initial velocity (u) = 0 m/s
Acceleration due to GRAVITY (g) = 10 m/s²

To FIND :

Time taken by the ball to touch the surface.
Final velocity of the ball.

SolutioN :

Use 2nd equation of motion :

$\implies \rm{s = ut + \dfrac{1}{2} gt^2} \\ \\ \\ \implies \rm{200 = 0 \times t + \bigg( \dfrac{1}{2} \times 10 \times t^2 \bigg) } \\ \\ \\ \implies \rm{200 = 0 + 5 t^2} \\ \\ \\ \implies \rm{5t^2 = 200} \\ \\ \\ \implies \rm{t^2 = \dfrac{200}{5}} \\ \\ \\ \implies \rm{t^2 = 40} \\ \\ \\ \implies \rm{t = \sqrt{40}} \\ \\ \\ \implies \rm{t = 6.324} \\ \\ \\ \large \implies {\boxed{\rm{t \: \approx \: 6.3 \: s}}}$

________________________

Now, use 1st equation of motion :

$\implies \rm{v = u + gt} \\ \\ \\ \implies \rm{v = 0 + 10 \times 6.3} \\ \\ \\ \large \implies {\boxed{\rm{v \: = \: 63 \: ms^{-1}}}}$