Question:medium

A body projected vertically up with an initial speed of $10\text{ ms}^{-1}$ reaches the point of projection after sometime with a speed of $8\text{ ms}^{-1}$. The maximum height reached by the body is (Acceleration due to gravity $= 10\text{ ms}^{-2}$)

Show Hint

When speed decreases upon return to the same point, mechanical energy is lost due to non-conservative forces like air drag. Use Work-Energy principle separately for ascent and descent.

Updated On: Mar 26, 2026

$5\text{ m}$
$3.2\text{ m}$
$4.1\text{ m}$
$4.5\text{ m}$

Show Solution

The Correct Option is C

Solution and Explanation

Was this answer helpful?

Questions Asked in TS EAMCET exam

If $\omega \neq 1$ is a cube root of unity, then one root among the $7^{th}$ roots of $(1+\omega)$ is

TS EAMCET - 2025
Complex numbers

View Solution

If $A = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \\ 1 & 3 & 1 \end{pmatrix}$ and $B = \begin{pmatrix} 2 & 3 & 4 \\ 3 & 2 & 2 \\ 2 & 4 & 2 \end{pmatrix}$, then $\sqrt{|\text{Adj}(AB)|} =$

TS EAMCET - 2025
Matrices and Determinants

View Solution

If $ p_1, p_2, p_3 $ are the altitudes and $ a=4, b=5, c=6 $ are the sides of a triangle ABC, then $ \frac{1}{p_1^2} + \frac{1}{p_2^2} + \frac{1}{p_3^2} = $

TS EAMCET - 2025
Geometry

View Solution

If $\alpha, \beta, \gamma, \delta, \epsilon$ are the roots of the equation $x^5 + x^4 - 13x^3 - 13x^2 + 36x + 36 = 0$ and $\alpha<\beta<\gamma<\delta<\epsilon$ then $ \frac{\epsilon}{\alpha} + \frac{\delta}{\beta} + \frac{1}{\gamma} = $

TS EAMCET - 2025
Algebra

View Solution

A body projected vertically up with an initial speed of $10\text{ ms}^{-1}$ reaches the point of projection after sometime with a speed of $8\text{ ms}^{-1}$. The maximum height reached by the body is (Acceleration due to gravity $= 10\text{ ms}^{-2}$)

Show Hint

The Correct Option is C

Solution and Explanation

Top Questions on Mechanics

Questions Asked in TS EAMCET exam