Question

Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 12 ​m/sec. Because the acceleration of gravity at the​ planet's surface was gs ​m/sec2​, the explorers expected the ball bearing to reach a height of s=12−​(1/2)gst2 ​m, t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gs​?

Answer 1

gs = 0.6 m/s^2

Step-by-step explanation:

Given data:

velocity = 12 m/s

height s = 12t -(1/2) g_s t^2

Given velocity is the derivatives of height

`v(t) = (d)/(dt) s(t)`

`= (d)/(dt)(12t -(1)/(2) g_s t^2)`

`= 12 - g_s t`

when velocity tend to 0 , maximum height is reached

`v(t) = 12 - g_s t`

`0 = 12 - g_s t`

`g_s = (12)/(t)`

at t = 20 sec ball reached the max height, so

`g_s = (12)/(20) = 0.6 m/s^2`