Question

A projectile is thrown from a point $O$ on the ground at an angle $45^{\circ}$ from the vertical and with a speed $5 \sqrt{2} \,m / s$. The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, $05\, s$ 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 $g =10 \,m / s ^{2}$. The value of $t$ is ____

Answer 1

Correct Answer: 0.5

Given:
- Launch angle from vertical = 45° ⇒ From horizontal = 45°
- Initial speed of projectile = 5√2 m/s
- Splits into two equal parts at the highest point
- One part falls vertically and takes 0.5 s to reach the ground
- Acceleration due to gravity, g = 10 m/s²

Step 1: Resolve the initial velocity
Since angle from vertical is 45°, from horizontal it is also 45°:
- Horizontal component of velocity = u_x = 5√2 × cos(45°) = 5√2 × (1/√2) = 5 m/s
- Vertical component of velocity = u_y = 5√2 × sin(45°) = 5 m/s

Step 2: Time to reach the highest point
At highest point, vertical velocity becomes zero.
Using: v = u − g·t
0 = 5 − 10·t ⇒ t = 0.5 s

So, the projectile reaches the highest point in 0.5 s.
At this point, the vertical velocity is 0, and the horizontal velocity is 5 m/s.

Step 3: Motion after splitting
At the highest point, the projectile splits into two equal parts:
- One part drops vertically down and hits the ground in 0.5 s ⇒ height h can be calculated:
Using: h = 0.5·g·t² = 0.5·10·(0.5)² = 1.25 m

Step 4: Second part of the projectile
It continues its motion horizontally from the same height (1.25 m) with horizontal velocity = 5 m/s
Let it take time t seconds to fall to the ground from 1.25 m:
Using: h = 0.5·g·t² ⇒ 1.25 = 0.5·10·t² ⇒ 1.25 = 5·t² ⇒ t² = 0.25 ⇒ t = 0.5 s

Final Answer:
The second part hits the ground after t = 0.5 seconds.