# A ball is thrown vertically upwards with a velocity of $49\ m/s$.Calculate$(i)$. the maximum height to which it rises,$(ii)$. the total time it takes to return to the surface of the earth.

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Given:
Initial velocity of the ball $u=49\ m/s$

The velocity of the ball at maximum height $v=0$

$g=9.8\ m/s^2$

To do:

$(i)$. To find the maximum height to which it rises,

$(ii)$. To find the total time it takes to return to the surface of the earth.

Solution:

Let $h$ be the maximum height to which it rises and $t$ be the time taken to reach the maximum height.

On using the third equation of the motion,

$v^2=u^2-2gh$      [take $g$ negative as it goes upward]

Or $2gh=v^2-u^2$

Or $2\times(-9.8)\times h=0-(49)^2$

Or $-19.6h=-2401$

Or $h=\frac{2401}{19.6}$

Or $h=122.5\ m$

Now consider a formula,

$v=u+gt$

Or $0=49+(-9.8)\times t$

Or $-49=-9.8t$

Or $t=\frac{49}{9.8}$

$t=5\ sec$

Therefore, the maximum height to which the ball rises $=122.5\ m$ and the total time ball takes to return to the surface of the earth $=5+5=10\ sec$.
Updated on 10-Oct-2022 13:22:44