Answer :
Certainly! Let's break down each part of the question step by step.
### (a) Height of Priyanshu
Priyanshu is walking away from the base of a lamp post at a speed of 1 m/s. After 2 seconds, his shadow is 1 meter long.
- Given Data:
- Lamp height ([tex]\( L \)[/tex]) = 4.5 meters
- Shadow length after 2 seconds ([tex]\( S_2 \)[/tex]) = 1 meter
- Priyanshu's walking speed ([tex]\( v \)[/tex]) = 1 m/s
- Situation after 2 seconds:
- Priyanshu's distance from the lamp post ([tex]\( D_2 \)[/tex]) = walking speed × time = 1 m/s × 2 s = 2 meters
Using the concept of similar triangles:
[tex]\[ \frac{L}{D_2 + S_2} = \frac{H}{S_2} \][/tex]
where [tex]\( H \)[/tex] is the height of Priyanshu.
Substitute the values:
[tex]\[ \frac{4.5}{2 + 1} = \frac{H}{1} \][/tex]
[tex]\[ \frac{4.5}{3} = H \][/tex]
[tex]\[ H = 1.5 \text{ meters} \][/tex]
So, the height of Priyanshu is 1.5 meters.
### (b) Minimum time after which Priyanshu's shadow becomes larger than his original height
Given Priyanshu's height ([tex]\( H \)[/tex]) = 1.5 meters, we need to find the time [tex]\( t \)[/tex] after which his shadow becomes longer than 1.5 meters.
From similar triangles:
[tex]\[ \frac{L}{D} = \frac{H}{S} \][/tex]
Reaching a time when shadow length [tex]\( S \)[/tex] > Priyanshu's height:
[tex]\[ S > H \][/tex]
Solving for time [tex]\( t \)[/tex]:
Distance from the lamp post at this moment:
[tex]\[ D = H \][/tex]
Plugging in the values:
[tex]\[ D = 1.5 \text{ meters} \][/tex]
Therefore, the time taken:
[tex]\[ t = \frac{D}{v} \][/tex]
[tex]\[ t = \frac{1.5}{1} \][/tex]
[tex]\[ t = 1.5 \text{ seconds} \][/tex]
So, the minimum time after which Priyanshu's shadow becomes larger than his height is 1.5 seconds.
### (c) Distance of Priyanshu from pole when his shadow becomes larger than his height
From the previous part, we found that the time [tex]\( t \)[/tex] is 1.5 seconds. Therefore, the distance from the pole at this point is:
[tex]\[ \text{Distance} = v \times t \][/tex]
[tex]\[ \text{Distance} = 1 \, \text{m/s} \times 1.5 \, \text{s} \][/tex]
[tex]\[ \text{Distance} = 1.5 \text{ meters} \][/tex]
So, the distance of Priyanshu from the pole when his shadow becomes larger than his height is 1.5 meters.
### (d) Length of shadow after 4 seconds
After 4 seconds, Priyanshu is:
[tex]\[ D_4 = v \times 4 \][/tex]
[tex]\[ D_4 = 1 \, \text{m/s} \times 4 \, \text{s} \][/tex]
[tex]\[ D_4 = 4 \text{ meters} \][/tex]
Using the concept of similar triangles:
[tex]\[ \frac{L}{D_4} = \frac{H}{S_4} \][/tex]
Rewriting the equation for the shadow length [tex]\( S_4 \)[/tex]:
[tex]\[ S_4 = \frac{D_4 \times H}{(D_4 - H)} \][/tex]
Substitute the known values:
[tex]\[ S_4 = \frac{4 \times 1.5}{(4 - 1.5)} \][/tex]
[tex]\[ S_4 = \frac{6}{2.5} \][/tex]
[tex]\[ S_4 = 2.4 \text{ meters} \][/tex]
So, the length of Priyanshu's shadow after 4 seconds is 2.4 meters.
### (a) Height of Priyanshu
Priyanshu is walking away from the base of a lamp post at a speed of 1 m/s. After 2 seconds, his shadow is 1 meter long.
- Given Data:
- Lamp height ([tex]\( L \)[/tex]) = 4.5 meters
- Shadow length after 2 seconds ([tex]\( S_2 \)[/tex]) = 1 meter
- Priyanshu's walking speed ([tex]\( v \)[/tex]) = 1 m/s
- Situation after 2 seconds:
- Priyanshu's distance from the lamp post ([tex]\( D_2 \)[/tex]) = walking speed × time = 1 m/s × 2 s = 2 meters
Using the concept of similar triangles:
[tex]\[ \frac{L}{D_2 + S_2} = \frac{H}{S_2} \][/tex]
where [tex]\( H \)[/tex] is the height of Priyanshu.
Substitute the values:
[tex]\[ \frac{4.5}{2 + 1} = \frac{H}{1} \][/tex]
[tex]\[ \frac{4.5}{3} = H \][/tex]
[tex]\[ H = 1.5 \text{ meters} \][/tex]
So, the height of Priyanshu is 1.5 meters.
### (b) Minimum time after which Priyanshu's shadow becomes larger than his original height
Given Priyanshu's height ([tex]\( H \)[/tex]) = 1.5 meters, we need to find the time [tex]\( t \)[/tex] after which his shadow becomes longer than 1.5 meters.
From similar triangles:
[tex]\[ \frac{L}{D} = \frac{H}{S} \][/tex]
Reaching a time when shadow length [tex]\( S \)[/tex] > Priyanshu's height:
[tex]\[ S > H \][/tex]
Solving for time [tex]\( t \)[/tex]:
Distance from the lamp post at this moment:
[tex]\[ D = H \][/tex]
Plugging in the values:
[tex]\[ D = 1.5 \text{ meters} \][/tex]
Therefore, the time taken:
[tex]\[ t = \frac{D}{v} \][/tex]
[tex]\[ t = \frac{1.5}{1} \][/tex]
[tex]\[ t = 1.5 \text{ seconds} \][/tex]
So, the minimum time after which Priyanshu's shadow becomes larger than his height is 1.5 seconds.
### (c) Distance of Priyanshu from pole when his shadow becomes larger than his height
From the previous part, we found that the time [tex]\( t \)[/tex] is 1.5 seconds. Therefore, the distance from the pole at this point is:
[tex]\[ \text{Distance} = v \times t \][/tex]
[tex]\[ \text{Distance} = 1 \, \text{m/s} \times 1.5 \, \text{s} \][/tex]
[tex]\[ \text{Distance} = 1.5 \text{ meters} \][/tex]
So, the distance of Priyanshu from the pole when his shadow becomes larger than his height is 1.5 meters.
### (d) Length of shadow after 4 seconds
After 4 seconds, Priyanshu is:
[tex]\[ D_4 = v \times 4 \][/tex]
[tex]\[ D_4 = 1 \, \text{m/s} \times 4 \, \text{s} \][/tex]
[tex]\[ D_4 = 4 \text{ meters} \][/tex]
Using the concept of similar triangles:
[tex]\[ \frac{L}{D_4} = \frac{H}{S_4} \][/tex]
Rewriting the equation for the shadow length [tex]\( S_4 \)[/tex]:
[tex]\[ S_4 = \frac{D_4 \times H}{(D_4 - H)} \][/tex]
Substitute the known values:
[tex]\[ S_4 = \frac{4 \times 1.5}{(4 - 1.5)} \][/tex]
[tex]\[ S_4 = \frac{6}{2.5} \][/tex]
[tex]\[ S_4 = 2.4 \text{ meters} \][/tex]
So, the length of Priyanshu's shadow after 4 seconds is 2.4 meters.
