Question

An object is thrown vertically upwards to a height of $10\mathrm{m}$.

i. Calculate its velocity with which it was thrown upward.

ii. Calculate its time taken by the object to reach the highest point.

Answer 1

Step 1. Given data:

Distance travelled, $\mathrm{s}=10\mathrm{m}$,

Final velocity, $\mathrm{v}=0\mathrm{m}/\mathrm{s}$

$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

We have to calculate the velocity with which it was thrown upward and the time taken by the object to reach the highest point.

Step 2. Formula to be used.

According to the third equation of motion,

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gs}$

According to the first equation of motion,

$\mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{g}}$

Here, $\mathrm{u}$ is initial velocity, $\mathrm{g}$ is the acceleration due to gravity, and $\mathrm{v}$ is final velocity and $\mathrm{t}$ is time.

Step 3. Calculate the velocity with which it was thrown upward.

So,

${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{gs}$

${\left(0\right)}^2-{\mathrm{u}}^2=2\times (-9.8)\times 10$

${\mathrm{u}}^2=196$

$\mathrm{u}=\sqrt{196}$

$\mathrm{u}=14\mathrm{m}/\mathrm{s}$

So, the velocity with which the object is thrown upwards is $14\mathrm{m}/\mathrm{s}$.

Step 4. Calculate the time taken by the object to reach the highest point.

The velocity with which the object is thrown upwards is $14\mathrm{m}/\mathrm{s}$.

So,

$\mathrm{t}=\frac{\mathrm{v}-\mathrm{u}}{\mathrm{g}}$

$\mathrm{t}=1.43\sec$

So, the time taken by the object to reach the highest point is $1.43\sec$.

Hence, the velocity with which the object is thrown upwards is $14\mathrm{m}/\mathrm{s}$ and the time taken by the object to reach the highest point is $1.43\sec$.