A ball is dropped from a height. If it takes 0.2s to cross the last 6m before hitting the ground, find the height from which it is dropped. Take g = 10m/ s²

Since ball takes 0.2 s to cross the last 6m before hitting the ground.

∴ Let velocity of the ball after time t₁ = u;

The figure below depicts the situation of the question.

By Newton’s second equation of motion;

Distance (S) = ut +

Here S = 6m; t = 0.2s; g = 10m/s²;

⇒ 6 = 0.2u +

⇒ 6 = 0.2u +

⇒ 6 = 0.2u + 0.2

⇒ 0.2u = 6 – 0.2

⇒ 0.2u = 5.8.

⇒ u = 29 m/s.

Hence final speed (v) of the ball after it traveled ‘X’ distance is 29m/s.

By Newton’s first equation of motion

⇒ v = u + at;

Here v = 29 m/s (final velocity).

u = 0 (Initial velocity is zero);

A = g = 10 m/s².

T = t₁ (refer figure above);

Putting the values we get

29 = 0 + 10 t₁;

⇒ t₁ =

Now finding the distance ‘X’ by newton’s second equation of motion

Distance (S) = ut +

Here S = X (refer figure); t = t₁ = 2.9s; g = 10m/s²; u = 0 (initial velocity);

Putting the values we get;

X = 0(2.9) +

⇒ X = 0 +

⇒ X = 5 x 8.41

⇒ X = 42.05 m

Now;

Total height from which ball was thrown = X + Y = 42.05 + 6

⇒ Total height = 48.05 m;