A body projected at a certain angle (neq 90^{ | vedclass

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

Question

(A) For a projectile motion, if a body passes through a point at time $t_1$ and reaches the ground after a further time $t_2$, the total time of fligh

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) For a projectile motion, if a body passes through a point at time $t_1$ and reaches the ground after a further time $t_2$, the total time of flight is $T = t_1 + t_2 = 2.3 \ s + 5.7 \ s = 8.0 \ s$.The time taken to reach the maximum height is $t_{max} = \frac{T}{2} = \frac{8.0 \ s}{2} = 4.0 \ s$.The maximum height $H$ is given by the formula $H = \frac{1}{2} g t_{max}^2$.Substituting the values: $H = \frac{1}{2} 	imes 10 \ ms^{-2} 	imes (4.0 \ s)^2$.$H = 5 	imes 16 = 80 \ m$.Thus, the maximum height reached by the body is $80 \ m$.

$A$ body projected at a certain angle $(\neq 90^{\circ})$ from the ground crosses a point in its path at a time $t_1 = 2.3 \ s$ and from there it reaches the ground after a further time $t_2 = 5.7 \ s$. The maximum height reached by the body is (Take $g = 10 \ ms^{-2}$) (in $m$)

Similar Questions

$A$ stone is projected at an angle of $30^{\circ}$ to the horizontal. The ratio of the kinetic energy of the stone at the point of projection to its kinetic energy at the highest point of its flight will be:

The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$

$A$ projectile crosses two walls of equal height $H$ symmetrically as shown. The velocity of projection is........ $ms^{-1}$.

In a projectile motion,the velocity at maximum height is:

$A$ particle is thrown with a speed $u$ at an angle $\theta$ with the horizontal. When the particle makes an angle $\phi$ with the horizontal,its speed changes to $v$,where

Vedclass Test Series

Exam Paper Generator

Online Exam Module