Question

On a different planet, a ball is fired straight upwards from the ground with an initial speed of 100 m/s. At the same instant, a second ball is dropped from the top of a cliff 400 m above the ground. The balls arrive at the same vertical location at the EXACT instant that the ball thrown upwards reaches its highest point. Determine the magnitude of the acceleration due to gravity on this planet. In other words, find g on this planet. You can assume there is no air resistance.

Answer 1

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

g = 25 m/s²