Question

A stone is thrown Vertically upward with an initial velocity of 50 m/s. The maximum height reached by the stone is: (g= 10 m/s2)​

Answer 1

☀️ Given that: A stone is thrown Vertically upward with an initial velocity of 50 m/s.

     Need to find: . The maximum height reached by the stone ( g = 10 m/s² )​

Must Know and Understand:

• When the stone will attain its maximum height the final velocity of the stone will be zero as it will come down after attaining maximum height and zero velocity.

• Acceleration acting upon the stone is gravitational acceleration  ( 10 m/s² approx ) which will be negative as the stone is thrown vertically upwards. [ Note - it will be positive if the stone is thrown downwards ]

• According to the third equation of motion v² = u² +  2as

Where,

v is final velocity ( 0 m/s )

u is initial velocity ( 50 m/s )

a is acceleration ( -10 m/s² )

s is distance covered

Solution:

Finding distance-

→ 0² = 50² + 2 × -10 × s

→ 0 = 2500 + ( -20s )

→ -2500 = -20s

→ s = -2500/-20

→ s = 2500/20

→ s = 250/2

→ s = 125m

⠀━━━━━━━━━━━━━━━━━━━  

• Henceforth, the maximum height reached by the stone is 125m