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 = 9.8 m/s²​)

Answer 1

Distance travelled by the ball = 19.6 m

Acceleration due to gravity = ​g = 9.8 m/s²​

Let the initial velocity of the body = u

Final velocity of the ball = 0 m/s

Time taken by the ball to reach to the highest point = t

Using the third equation of motion, $v^2=u^2-2gs$

$$\begin{gathered} \Rightarrow 0=u^2-2g\times 19.6 \\ \Rightarrow u^2=2\times 9.8\times 19.6=(19.6{)}^2 \\ \Rightarrow u=19.6\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, using the first equation of motion for the ball thrown upward, v = u - gt

$$\begin{gathered} \therefore 0=u-gt \\ \Rightarrow t=\frac{u}{g}=\frac{19.6\mathrm{m}/\mathrm{s}}{9.8\mathrm{m}/{\mathrm{s}}^2}=2\mathrm{s} \\ \Rightarrow t=2\mathrm{s} \end{gathered}$$


Time taken by the ball to reach the highest point = 2 s

And the initial velocity of the ball, u = 19.6 m/s