A projectile is thrown from a point $\mathrm{O}$ on the ground at an angle $45^{\circ}$ from the vertical and with a speed $5 \sqrt{2} \mathrm{~m} / \mathrm{s}$. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, $0.5…

Question

A projectile is thrown from a point $\mathrm{O}$ on the ground at an angle $45^{\circ}$ from the vertical and with a speed $5 \sqrt{2} \mathrm{~m} / \mathrm{s}$. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, $0.5 \mathrm{~s}$ after the splitting. The other part, $t$ seconds after the splitting, falls to the ground at a distance $x$ meters from the point O. The acceleration due to gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$.
The value of

t

is _________.

Aicharya · Accepted Answer

After splitting of projectile in two equal parts one part falls vertically down to the ground.

So, velocity of another part will be according to conservation of linear momentum,

$m \times 5 = \frac{m}{2} \times 0 + \frac{m}{2} \times v$

$\Rightarrow v = 10 m s^{- 1}$

Now time taken by $m_{1}$ , to reach ground,

since after splitting of projectile $m_{1}$ will perform horizontal projectile so,

$t = \sqrt{\frac{2 h}{g}}$

$\Rightarrow t = \sqrt{\frac{2 u^{2} \sin^{2} θ}{2 g^{2}}}$

$\Rightarrow t = \sqrt{\frac{2 \times 50}{2 \times 10 \times 10} \times \frac{1}{2}} = \sqrt{0.25}$

$\Rightarrow t = 0.5 s$

A projectile is thrown from a point $\mathrm{O}$ on the ground at an angle $45^{\circ}$ from the vertical…

Solution