Question

1) A stone is thrown vertically upwards with a speed of 20mls from the top of a building.the hieght of building is 25.0m from the ground.
a) how high will the ball rise?
b) how long will it take to reach the ground?(g=10m/s)​

Answer 1

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(a) Here,

u = + 20 m/s

g = - 10 m/s²

At the highest point, v = 0

Suppose the balls rise to the highest h from the point of projection.

v² - u² = 2gs

⇒ (0)² - (20)² = 2 × (- 10) × h

⇒ h = + 20 m.

Hence, the ball rises to 20 m.

(ii) Net displacement, s = - 25 m.

Negative sign is taken because displacement is in the opposite direction of the initial velocity.

s = ut + 12 gt²

⇒ - 25 = 20t + 1/2 × (- 10) × t²

⇒ 5t² - 20t - 25 = 0

⇒ t² - 4t - 5 = 0

⇒ (t + 1) (t - 5) = 0

⇒ t ≠ - 1, 5 seconds

⇒ t = 5 seconds.

Hence, the ball hits the ground to 5 seconds.

Answer 2

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(a) Here,

u = + 20 m/s

g = - 10 m/s²

At the highest point, v = 0

Suppose the balls rise to the highest h from the point of projection.

v² - u² = 2gs

⇒ (0)² - (20)² = 2 × (- 10) × h

⇒ h = + 20 m.

Hence, the ball rises to 20 m.

(ii) Net displacement, s = - 25 m.

Negative sign is taken because displacement is in the opposite direction of the initial velocity.

s = ut + 12 gt²

⇒ - 25 = 20t + 1/2 × (- 10) × t²

⇒ 5t² - 20t - 25 = 0

⇒ t² - 4t - 5 = 0

⇒ (t + 1) (t - 5) = 0

⇒ t ≠ - 1, 5 seconds

⇒ t = 5 seconds.

Hence, the ball hits the ground to 5 seconds.

HOPE SO IT WILL HELP......