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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.
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
14K+ Views
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