Question

an object thrown vertically upwards comes back after 6 second. calculate-1. the velocity with which it was thrown. 2. the maximum height it reached?​

Answer 1

$\huge\mathfrak\red{Answer :}$

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$<b>Given,$

Initial Velocity = 0 [ Body Starts From Rest ]

Gravitational Acceleration = 9.8 ms^-2

Time Taken = 6/2 = 3 seconds

From First Equation of Motion

» v = u + at

» v = 0 + 9.8 × 3

» v = 29.4 ms^-1

★ Hence, Velocity Will Be 29.4 ms^-1 or 30 ms^-1

From Second Equation of Motion

» S = ut + 1/2 at^2

» S = 30*3 + 1/2 × (-9.8) × (3)^2

» 90 + (-4.9) × 9

» 90 - 44.1

» 45.9 m.

★ Hence, the Distance Covered Will be 46m.

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