Question

When a body is thrown vertically up wi
th a velocity 19.6m/s find the maximum height reached,time of ascent

Answer 1

Answer:

The Maximum height reached = 19.6 m, Time of ascent = 2 sec

Explanation:

At the maximum height the velocity becomes zero.

The gravitational acceleration acts as negative acceleration as the object is moving upward against the gravitational force.

Therefore:

Initial Velocity = u = 19.6m/s

Final Velocity = 0

Acceleration = - 9.8$m/sec^2$

$v^2 = u^2+2as\\v^2-u^2=2as\\\frac{v^2-u^2}{2a}=s\\\frac{0^2-19.6^2}{2*(-9.8)}=s\\\frac{-19.6*19.6}{-2*9.8}=s\\19.6m=s\\Displacement = 19.6m$

The maximum height = 19.6 meter

$v=u+at\\v-u=at\\\frac{v-u}{a}=t\\\frac{0-19.6}{-9.8}=t\\\frac{-19.6}{-9.8}=t\\2=t$

Time of ascent = 2 sec