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

Answer:

gs = 0.6 m/s^2

Explanation:

Given data:

velocity = 12 m/s

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

Given velocity is the derivatives of  height

$v(t) = \frac{d}{dt} s(t)$

      $= \frac{d}{dt}(12t -\frac{1}{2} g_s t^2)$

      $= 12 - g_s t$

when velocity tend to 0 , maximum height is reached

$v(t) = 12 - g_s t$

$0 = 12 - g_s t$

$g_s = \frac{12}{t}$

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

$g_s = \frac{12}{20} = 0.6 m/s^2$