Question

A body is thrown vertically upward, such that the distance travelled by it in third and sixth seconds are equal. The maximum height reached by the body is-

Answer 1

8g

Answer 2

The displacement of a body thrown vertically upward in the $n^{th}$ second is given by the formula: $d_n = u + a(n - 1/2)$ where $u$ is the initial velocity and $a$ is the acceleration. For motion under gravity, $a = -g$. So, the displacement in the $n^{th}$ second is: $d_n = u - g(n - 1/2)$

The distance travelled ($S_n$) in the $n^{th}$ second is equal to the magnitude of the displacement ($|d_n|$) if the body's velocity does not change direction during that second.

For the 3rd second ($n=3$): $d_3 = u - g(3 - 1/2) = u - 2.5g$

For the 6th second ($n=6$): $d_6 = u - g(6 - 1/2) = u - 5.5g$

The problem states that the distance travelled in the 3rd and 6th seconds are equal: $S_3 = S_6$. Let $t_{up} = u/g$ be the time to reach the maximum height.

If the body moves upwards throughout both intervals ($t_{up} \ge 6$), then $S_3 = d_3$ and $S_6 = d_6$. $u - 2.5g = u - 5.5g \implies 2.5g = 5.5g$, which is impossible.

If the body moves downwards throughout both intervals ($t_{up} \le 2$), then $S_3 = |d_3|$ and $S_6 = |d_6|$. $|u - 2.5g| = |u - 5.5g|$. This equation implies either $u - 2.5g = u - 5.5g$ (impossible) or $u - 2.5g = -(u - 5.5g)$. $u - 2.5g = -u + 5.5g$ $2u = 8g \implies u = 4g$. However, this contradicts the condition $t_{up} \le 2$ (since $u=4g$ implies $t_{up}=4$). So, this case is invalid.

The only way for $S_3 = S_6$ to hold is if the magnitudes of displacements are equal, i.e., $|d_3| = |d_6|$. This occurs when $u=4g$. If $u = 4g$, then $t_{up} = u/g = 4$ seconds.

• 3rd second ($t=2$ to $t=3$): Since $t_{up}=4s$, the body is still moving upwards during this interval. So, distance travelled $S_3 = d_3 = u - 2.5g = 4g - 2.5g = 1.5g$.

• 6th second ($t=5$ to $t=6$): Since $t_{up}=4s$, the body is moving downwards during this interval. The displacement is $d_6 = u - 5.5g = 4g - 5.5g = -1.5g$. The distance travelled $S_6 = |d_6| = |-1.5g| = 1.5g$.

Thus, $S_3 = S_6 = 1.5g$, satisfying the condition. The initial velocity is $u = 4g$.

The maximum height ($H$) reached by the body is given by: $H = \frac{u^2}{2g}$ Substituting $u = 4g$: $H = \frac{(4g)^2}{2g} = \frac{16g^2}{2g} = 8g$