Question

The value of t is ________.

Answer 1

Correct Answer: 0.5

Step 1: Understanding the problem setup
A projectile is thrown at an angle of \( 45^{\circ} \) from the vertical with an initial speed of \( 5 \sqrt{2} \, \text{m/s} \). The projectile splits into two equal parts at the highest point of its trajectory. One part falls vertically to the ground 0.5 seconds after the splitting.
The other part lands at a distance \( x \) meters from the point of projection after \( t \) seconds.
The acceleration due to gravity is given as \( g = 10 \, \text{m/s}^2 \).
Step 2: Analyzing the motion of the projectile before splitting
The projectile is thrown with a velocity of \( 5 \sqrt{2} \, \text{m/s} \) at an angle of \( 45^{\circ} \) from the vertical. The components of the initial velocity are:
- Vertical velocity \( v_y = 5 \sqrt{2} \, \cos(45^{\circ}) = 5 \, \text{m/s} \)
- Horizontal velocity \( v_x = 5 \sqrt{2} \, \sin(45^{\circ}) = 5 \, \text{m/s} \)
At the highest point of the trajectory, the vertical velocity becomes zero. Therefore, the projectile’s time to reach the highest point can be calculated using the formula for the time of flight under constant acceleration:
\[ t_{\text{up}} = \frac{v_y}{g} = \frac{5}{10} = 0.5 \, \text{seconds} \] So, it takes 0.5 seconds to reach the highest point.
Step 3: Analyzing the motion of the parts after the splitting
When the projectile splits at the highest point, one part falls vertically down. The time it takes to fall to the ground is given as 0.5 seconds. The second part falls at a distance \( x \) meters from the point of origin after \( t \) seconds.
The second part of the projectile continues to move horizontally with the initial horizontal velocity \( v_x = 5 \, \text{m/s} \). The time for it to fall to the ground is also \( t \), and during this time, it moves horizontally a distance of \( x = v_x \cdot t \). Since the part falls from the highest point, the time taken for it to reach the ground is \( t = 0.5 \, \text{seconds} \).
Therefore, the time \( t \) taken by the second part to land at a distance \( x \) from the point of origin is:
\[ t = 0.5 \, \text{seconds} \] Step 4: Conclusion
Hence, the value of \( t \) is \( 0.50 \, \text{seconds} \).

Answer 2

Step 1: Understanding the given data
We are given the following data:
- The projectile is thrown at an angle of \( 45^{\circ} \) from the vertical with an initial speed of \( 5\sqrt{2} \, \text{m/s} \).
- At the highest point of its trajectory, the projectile splits into two equal parts.
- One part falls vertically down to the ground, 0.5 seconds after the splitting.
- The other part, denoted as \( t \), falls to the ground at a distance \( x \) meters from the point \( O \).
- The acceleration due to gravity is \( g = 10 \, \text{m/s}^2 \).
Step 2: Analyzing the projectile motion
Since the projectile is thrown at an angle of \( 45^{\circ} \) from the vertical, the initial velocity components can be found using the angle \( \theta = 45^{\circ} \):
- Initial velocity in the vertical direction: \( u_y = 5 \sqrt{2} \times \cos(45^{\circ}) = 5 \, \text{m/s} \)
- Initial velocity in the horizontal direction: \( u_x = 5 \sqrt{2} \times \sin(45^{\circ}) = 5 \, \text{m/s} \)
The total time to reach the highest point is when the vertical velocity becomes zero. The time to reach the highest point \( t_{\text{up}} \) can be found from the equation for vertical motion:
\[ v_y = u_y - g t_{\text{up}} \] At the highest point, \( v_y = 0 \), so:
\[ 0 = 5 - 10 t_{\text{up}} \quad \Rightarrow \quad t_{\text{up}} = 0.5 \, \text{seconds} \] Therefore, the total time to reach the highest point is 0.5 seconds.
Step 3: Analyzing the motion of the two parts after splitting
After splitting, one part falls vertically downward. Since it takes 0.5 seconds to reach the highest point, the time for the vertically falling part to hit the ground is 0.5 seconds.
The other part falls at a distance \( x \) from the point \( O \). The horizontal velocity of this part is the same as the initial horizontal velocity of the projectile, which is \( 5 \, \text{m/s} \).
The horizontal distance traveled by this part is given by:
\[ x = u_x \times t \] where \( u_x = 5 \, \text{m/s} \) and \( t \) is the time it takes for the part to fall to the ground.
Since the time for the vertically falling part to hit the ground is 0.5 seconds, the time for the second part to reach the ground is also 0.5 seconds (since it is falling from the highest point). Therefore:
\[ x = 5 \times 0.5 = 7.5 \, \text{m} \] Step 4: Conclusion
The value of \( x \) is \( \boxed{7.5} \, \text{meters} \).