From the top of a tower,a ball is thrown vert | vedclass

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

Question

(A) Let the height of the tower be $h$ and the speed of projection in the first two cases be $u$. Taking the downward direction as positive.For case-$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let the height of the tower be $h$ and the speed of projection in the first two cases be $u$. Taking the downward direction as positive.For case-$I$ (thrown upward): The displacement is $h$. The initial velocity is $-u$. The acceleration is $g$.Using the equation $s = ut + \frac{1}{2}at^2$:$h = -u(6) + \frac{1}{2}g(6)^2$$h = -6u + 18g$ ... $(i)$For case-$II$ (thrown downward): The displacement is $h$. The initial velocity is $u$. The acceleration is $g$.$h = u(1.5) + \frac{1}{2}g(1.5)^2$$h = 1.5u + 1.125g$ ... $(ii)$To eliminate $u$,multiply equation $(ii)$ by $4$:$4h = 6u + 4.5g$ ... $(iii)$Adding equation $(i)$ and equation $(iii)$:$h + 4h = (-6u + 18g) + (6u + 4.5g)$$5h = 22.5g$$h = 4.5g$ ... $(iv)$For case-$III$ (released from rest): The initial velocity is $0$. The acceleration is $g$.$h = 0(t) + \frac{1}{2}gt^2$$h = \frac{1}{2}gt^2$ ... $(v)$Substituting equation $(iv)$ into equation $(v)$:$4.5g = \frac{1}{2}gt^2$$t^2 = 9$$t = 3 \, s$.

Similar Questions

The ratio of the displacements of a freely falling body during the first, second, and third seconds of its motion is

$A$ balloon rises from rest with a constant acceleration $g/8$. $A$ stone is released from it when it has risen to height $h$. The time taken by the stone to reach the ground is

$A$ ball is thrown vertically upwards from the top of a tower at $4.9 \ m/s$. It strikes the pond near the base of the tower after $3 \ s$. The height of the tower is........$m$

$A$ body is dropped from a height $h$. It covers a distance of $9h/25$ in the last second of its motion. What is the height $h$ in $m$?

From a tower of height $H$, a particle is thrown vertically upwards with a speed $u$. The time taken by the particle to hit the ground is $n$ times that taken by it to reach the highest point of its path. The relation between $H, u$ and $n$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module