Question

A particle is thrown vertically upwards. Its velocity at half of the height is 10m/s, then the maximum height attained by it will be: (g = 10 m/s²).

• A. 10 m
• B. 20 m
• C. 15 m
• D. 25 m

Answer 1

The correct option is A

10 m

Step (1) Given:

The velocity at the half of the height, $v=10m/s$

The acceleration due to gravity, $g=10m/s^2$

Step (2) Finding the equation of motion when the particle is at the half of the height:

Let the maximum height be $h$ meter.

By the equation of motion formulas, we know

$v^2=u^2–2gs$

where $v$ is final velocity, $g$ is the acceleration due to gravity and $s$ is displacement.

$$\begin{gathered} {10}^2=u^2–2g\frac{h}{2} \\ 100=u^2–gh\dots \left(i\right) \end{gathered}$$

Step (3) Finding the equation of motion when the particle is at the maximum height:

At the maximum height, $v=0$

$$\begin{gathered} 0=u^2–2gh \\ {u}^2=2gh\dots ..\left(ii\right) \end{gathered}$$

Step (4) Finding the maximum height using the equations (i) & (ii):

Putting the value of u² in equation (i) we get

$$\begin{gathered} 100=2gh–gh \\ 100=gh \\ h=\frac{100}{g} \\ =\frac{100}{10} \\ =10m \end{gathered}$$

Hence, the answer is (A) 10 m