A particle is projected up with initial speed $u=10 \mathrm{~m} / \mathrm{s}$ from the top of a building at time $t=0$. At time $t=5 \mathrm{sec}$, the particle strikes the ground. Find the height/15 of the building.

Question

Aicharya · Accepted Answer

The particle is "freely falling" always. The expression "freely falling" does not necessarily mean an object is falling down A freely falling object is any object moving either upward downward under the influence of gravity alone.
The particle starts from $A$ and finally reaches at $C$.
Let us take origin at $A$. Upward direction is taken as positive and downward direction is taken as negative.
The particle moves in gravitational field where acceleration due to the gravity is always acting in downward direction whether it is moving upward or downward.
Hence acceleration vector $\vec{a}$ will always be $-10 \mathrm{~m} / \mathrm{s}^{2}, \mathfrak{s}$ its magnitude as well as direction remains constant always throughout the motion.

Hence acceleration, $\vec{a}=-10 \mathrm{~m} / \mathrm{s}^{2}$
Initially at $t=0$, the particle is projected in upward direction. Hence, initial velocity $\vec{u}=10 \mathrm{~m} / \mathrm{s}^{2}$
The particle moves from $A$ to $B$ (upward) and then $B$ to $C$ (downward).
The motion of the particle goes from $A$ to $B$ and then again passes point $A$. The net displacement of the particle upto this instant is zero. Then particle crosses point A and finally reaches to C. We know that net displacement is equal to the difference of fol position vector.
Hence net displacement of the particle during motion $(t=5 \mathrm{sec})$
Using $\vec{s}=\vec{u} t+\frac{1}{2} \vec{a} t^{2}$
$\Rightarrow \quad \vec{s}=(10) \times 5+\frac{1}{2}(-10)(5)^{2}=50-125=75(\mathrm{~m})$
Hence $H=75 \mathrm{~m}$

A particle is projected up with initial speed \(u=10 \mathrm{~m} / \mathrm{s}\) from the top of a building…

Solution