Question

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 1

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 :

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

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Now, use 1st equation of motion :

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

Answer 2

Answer :

Time = 6.3 seconds

Velocity = 63 m/s

Explanation :

Use 2nd equation of motion :

→ s = ut + 1/2 at²

→ 200 = 0 * t + 1/2 * 10 * t²

→ 200 = 0 + 5 t²

→ 5t² = 200

→ t² = 40

→ t = 6.3 s

$\rule{200}{2}$

Now, use 1st equation of motion :

→ v = u + at

→ v = 0 + 10 * 6.3

→ v = 63 m/s