Question

When a ball is thrown vertically upwards, it goes through a distance of 19.6 m. Find the initial velocity of the ball and the time taken by it to rise to the highest point. (Acceleration due to gravity, $g=9.8m{s}^{-2}$) [3 MARKS]

Answer 1

Problem: 3 Marks
Given, distance, $s=19.6m$, acceleration due to gravity, $g=-9.8m{s}^{-2}$ (negative sign is taken because ball is thrown up against gravity) and final velocity, $v=0$.
Let $u$ be the initial velocity and $t$ be the time taken to reach the highest point.
From third equation of motion,
$v^2=u^2+2gs$
$\Rightarrow (0{)}^2=u^2+2\times (-9.8)\times 19.6$
$\Rightarrow u^2=(19.6{)}^2$
$\Rightarrow u=19.6m{s}^{-1}$
Thus, the initial velocity of the ball is $19.6m{s}^{-1}$

From first equation of motion,
$v=u+at\Rightarrow 0=19.6+(-9.8\times t)$
$\Rightarrow \frac{19.6}{9.8}$
$\Rightarrow t=2s$

Thus, the ball takes 2 seconds to reach the highest point of its upward journey. Please note that the ball will take an equal time, that is, 2 seconds to fall to the ground. In other words, the ball will take a total of 2 + 2 = 4 seconds to reach back to the thrower.