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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.
$178.5\, m$ , $ 14\, s$
$155.5\, m$ , $ 16\, s$
$143.5\, m$ , $ 15\, s$
$122.5\, m$ , $ 10\, s$
Solution
According to the equation of motion under gravity $v^2 – u^2 = 2gs$
Where,
$u =$ Initial velocity of the ball
$v =$ Final velocity of the ball
$ s =$ Height achieved by the ball
$g =$ Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., $v = 0 \,m/s$ and $u = 49\, m/s$
During upward motion, $g = – 9.8\, m \,s^{-2}$
Let $h$ be the maximum height attained by the ball.
Hence, using $v^{2}-u^{2}=2 g s$
We have, $0^{2}-49^{2}=2(-9.8) h \Rightarrow h=\frac{49 \times 49}{2 \times 9.8}=122.5\, m$
Let $t$ be the time taken by the ball to reach the height $122.5 \,m$, then according to the equation of motion $v=u+g t$
We get,
$0=49+(-9.8) t \Rightarrow 9.8 t=49 \Rightarrow t=\frac{49}{9.8}=5\, s$
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return $= 5 + 5 = 10\, s$
