Question

a stone is thrown vertically upwards against gravity with the velocity of 20 metre per second.calculate the maximum height reached by the stone.​

Answer 1

Given

• Initial velocity = 20 m/s
• g = 10 m/s²

To Find

• Maximum height

Solution

☯ H = u²sin²θ/2g

• Here we are using the H_max equation from projectile motion

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✭ According to the Question :

➞ H = u²sin²θ/2g

• u = Initial velocity = 20 m/s
• θ = Angle of projectile = 90°
• g = Gravitational acceleration = 10 m/s²

➞ H = (20²×1)/2(10)

➞ H = 400/20

➞ H_max = 20 m

∴ The maximum height reached by the stone is 20 m

Answer 2

Explanation:

topic :

• velocity

given :

• a stone is thrown vertically upwards against gravity with the velocity of 20 metre per second.calculate the maximum height reached by the stone.

to find :

• calculate the maximum height reached by the stone.

solution :

• Well, if we use the conservation of energy, all the kinetic energy that it has at the bottom is equal to the potential energy it will have at the top. Quick and dirty assumption that g = 10 m/s²

• U=2m/s

• V=0

• A=-10m/s^2

• S= ?

• Now use equation third

• V^2= u^2+2as

• 0^2=20^2+2×-10×s

• 0=400+2×-10×s

• 0–400=-20×s

• -400/-20=s

• 20= s

• S= 20m

answer:

• the answer is S= 20 m