Question

A stone is dropped from a height 300 m and at the same time another stone is projected vertically
upwards with a velocity of 100m/sec. Find when & Where the two stones meet ?​

Answer 1

Answer:

height covered by the projectile from the ground

h=ut−

2

1

gt

2

h=100t−

2

1

×9.8×t

2

h=100t–4.9t

2

…(1)

Distance covered by the projectile thrown down in time t will be, since it falls down from the restu=0

300–h=

2

1

gt

2

300–(100t–4.9t

2

)=4.9t

2

…byequation(1)

300=100t,t=3s

Thus particles meet each other at 3s

Height at which particles meet

h=100t–4.9t

2

h=100×3–4.9×9

h=255.9m

Explanation:

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Answer 2

Here, it's given that

• Height (s) is 300 m
• Acceleration due to gravity (g) is -10 N (it's -9.8 N but for simplifying we take -10 N)
• Initial Velocity (u) = 100 m/s
• Final Velocity (v) = 0 ms

So, let the time required here be t and the height be x (where they meet)

∴ x = ut + ½gt²

⟹ x = 100 × t + ½ × -10× t²

⟹ x = 100t + ½ × -10t²

⟹ x = 100t - 5t²

Now,

300 - x = ut + ½gt²

Here,

• Initial Velocity (u) = 0 m/s
• Acceleration due to gravity (g) = 10 N

⟶ 300 - x = 0 × t + ½ × 10× t²

⟶ 300 - x = 0 + 5t²

⟶ 300 - x = 5t²

⟶ x = 300 - 5t²

Now, the Equation

⇢ 100t - 5t² = 300 - 5t²

⇢ 100 t = 300

⇢ t = 300/100

⇢ t = 3

Therefore, time = 3 seconds

Now, we can find the Value of x

⟹ 300 - 5t²

⟹ 300 - 5 ×3²

⟹ 300 - 5 × 9

⟹ 300 - 45

⟹ 255

Therefore, time at which the two stones meet is after 3 seconds and the distance is 255 metre.