A particle is projected vertically upwards fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $u$ be the initial velocity of the particle at point $A$.The total time taken to reach the ground again is $T = t_1 + t_2$.Using the equation

Vedclass · Accepted Answer

$\frac{1}{2} g t_1 t_2$

Vedclass · Answer

(D) Let $u$ be the initial velocity of the particle at point $A$.The total time taken to reach the ground again is $T = t_1 + t_2$.Using the equation of motion for the total flight: $0 = u(t_1 + t_2) - \frac{1}{2} g(t_1 + t_2)^2$.Solving for $u$: $u = \frac{1}{2} g(t_1 + t_2)$.The height $h$ of point $B$ from the ground is given by the displacement at time $t_1$: $h = u t_1 - \frac{1}{2} g t_1^2$.Substituting the value of $u$: $h = \left[ \frac{1}{2} g(t_1 + t_2) \right] t_1 - \frac{1}{2} g t_1^2$.Simplifying the expression: $h = \frac{1}{2} g t_1^2 + \frac{1}{2} g t_1 t_2 - \frac{1}{2} g t_1^2$.Therefore,$h = \frac{1}{2} g t_1 t_2$.

$A$ particle is projected vertically upwards from a point $A$ on the ground. It takes time $t_1$ to reach a point $B$,and it continues to move up. If it takes a further time $t_2$ to reach the ground from point $B$,then the height of point $B$ from the ground is:

Similar Questions

The number of values of $x$ in the interval $[0, 3\pi]$ satisfying the equation $2 \sin^2 x + 5 \sin x - 3 = 0$ is

If $f$ is a differentiable function such that $f(2x+1) = f(1-2x)$ for all $x \in R$,then find the minimum number of roots of the equation $f'(x) = 0$ in the interval $x \in (-5, 10)$,given that $f(2) = f(5) = f(10)$.

Which of the following compounds is optically active?

If the charge on a body is increased by $2 \ C$,the energy stored in it increases by $21 \%$. The original charge on the body in coulomb is

$A$ stunt man jumps his car over a crater as shown (neglect air resistance).

Vedclass Test Series

Exam Paper Generator

Online Exam Module