Aboy standing on the top of a cliff projects an object upwards with a velocity of 60m/s.
after 10s he releases anothe object from the cliff so that the 2 obects meet at a specific point. find the distance from the top of the cliff to the point these objects meet with the correct  explanation .
  (ans. 31.25m)
i need in this

Explanation:

For projectile motion, use constant acceleration equation:

Δx = v₀ t + ½ at²

where Δx is the displacement,

v₀ is the initial velocity,

a is the acceleration,

and t is time.

For the first object:

Δx = 60 t + ½ (-10) t²

Δx = 60 t − 5 t²

For the second object:

Δx = 0 (t−10) + ½ (-10) (t−10)²

Δx = -5 (t−10)²

When they meet, they have the same displacement, so:

60 t − 5 t² = -5 (t−10)²

60 t − 5 t² = -5 (t² − 20t + 100)

60 t − 5 t² = -5 t² + 100 t − 500

60 t = 100 t − 500

40 t = 500

t = 12.5

Plug into either of the original equations to find the displacement.

Δx = -5 (t−10)²

Δx = -5 (12.5−10)²

Δx = -31.25

The distance from the top of the cliff to the point where the objects meet is 31.25 meters.

the answer is b ) square root of 5

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