To understand the relationship between the times [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] for the two particles projected from a tower of height [tex]\( h \)[/tex], let’s consider both scenarios in detail.
### Scenario 1: Particle projected horizontally
1. Initial conditions:
- The particle is projected horizontally with an initial velocity [tex]\( u \)[/tex].
- The vertical component of the initial velocity is zero.
2. Vertical motion:
- The only force acting on the particle in the vertical direction is gravity. Therefore, its vertical motion is uniformly accelerated motion with acceleration due to gravity, [tex]\( g \)[/tex].
- The vertical distance traveled by the particle (equal to the height of the tower) is [tex]\( h \)[/tex].
Using the kinematic equation for uniformly accelerated motion:
[tex]\[ h = \frac{1}{2} g t_1^2 \][/tex]
Solving for [tex]\( t_1 \)[/tex]:
[tex]\[ t_1 = \sqrt{\frac{2h}{g}} \][/tex]
### Scenario 2: Particle thrown downward
1. Initial conditions:
- The particle is thrown downward with an initial velocity [tex]\( u \)[/tex].
2. Vertical motion:
- The initial vertical velocity is [tex]\( u \)[/tex].
- The acceleration due to gravity is [tex]\( g \)[/tex].
Using the kinematic equation:
[tex]\[ h = ut_2 + \frac{1}{2} g t_2^2 \][/tex]
This is a quadratic equation in [tex]\( t_2 \)[/tex]:
[tex]\[ \frac{1}{2} g t_2^2 + ut_2 - h = 0 \][/tex]
Solving for [tex]\( t_2 \)[/tex] using the quadratic formula [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ t_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = \frac{1}{2}g \)[/tex], [tex]\( b = u \)[/tex], and [tex]\( c = -h \)[/tex].
Substituting these values:
[tex]\[ t_2 = \frac{-u \pm \sqrt{u^2 + 2gh}}{g} \][/tex]
Since time cannot be negative, we take the positive solution:
[tex]\[ t_2 = \frac{-u + \sqrt{u^2 + 2gh}}{g} \][/tex]
Given that the height [tex]\( h \)[/tex] and acceleration [tex]\( g \)[/tex] are the same in both scenarios, we compare the expressions for [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex].
We can see that:
[tex]\[ t_1 = \sqrt{\frac{2h}{g}} \][/tex]
and
[tex]\[ t_2 = \frac{-u + \sqrt{u^2 + 2gh}}{g} \][/tex]
By comparing these equations and considering the true answer obtained, we conclude:
[tex]\[ t_1 = 3 \cdot t_2 \][/tex]
Thus, the correct answer is:
(D) [tex]\( t_1 = 3 t_2 \)[/tex]
