Answer:
g = 25 m/s²
Explanation:
Since, the balls meet at the exact instant when upward thrown ball reaches its maximum point. Therefore, applying 1st equation of motion to it, we get:
Vf₁ = Vi₁ + gt
where,
g = -g (for upward motion) = acceleration due to gravity at that planet = ?
t = time
Vf₁ = Final Velocity of Upward Thrown ball = 0 m/s (ball stops at highest point)
Vi₁ = Initial Velocity of Upward Thrown Ball = 100 m/s
Therefore,
0 m/s = 100 m/s - gt
gt = 100 m/s ------------- equation 1
Now, applying 3rd equation of motion for the height covered:
2(-g)h₁ = Vf₁² - Vi₁²
h₁ = 10000/2g
Now, we apply 2nd equation of motion to second ball moving downward:
h₂ = Vi₂t + (0.5)gt²
where,
h₂ = height covered by second ball at the time of meeting
Vi₂ = initial velocity of second ball = 0 m/s (since, it starts from rest)
Therefore,
h₂ = (0)(t) + (0.5)gt²
h₂ = (0.5)gt²
Now, it is clear from the given condition, that when the two balls meet, the sum of distance covered by both the balls will be equal to 400 m. Therefore,
h₁ + h₂ = 400 m
using values:
10000/2g + (0.5)gt² = 400
10000 + g²t² = (400)(2g)
using equation 1:
10000 + (100)² = 800g
g = 20000/800
<u>g = 25 m/s²</u>
