Question

Explorers on a planet with a thin atmosphere want to measure the acceleration of gravity at its surface. So they use a spring gun to launch a ball bearing vertically upward from the surface with a launch velocity of 29 ms. If we denote by gs the acceleration of gravity in that planet, in ms2, then the ball bearing will move following the formula:

s(t) = 30(t) - [gs/2][t^2]

If the ball bearing reaches its maximum height 21 seconds after being launched, what is the value of gs?

Answer 1

Answer:

The value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$.

Step-by-step explanation:

We know that the position function is given by

$s(t)=30t-\frac{g_s}{2} t^2$

Velocity is defined as the rate of change of position. Therefore,

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

So, the velocity function is

$v(t)=\frac{d}{dt}(s(t))\\\\v(t)=\frac{d}{dt}(30t-\frac{g_s}{2} t^2)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(30t\right)-\frac{d}{dt}\left(\frac{g_s}{2}t^2\right)\\\\v(t)=30-g_st$

When a body reaches a vertical velocity of zero, this is the maximum height of the body and then gravity will take over and accelerate the object downward. Thus,

$v(t)=30-g_st=0\\\\g_s=\frac{30}{t}$

We know that the ball bearing reaches its maximum height 21 seconds after being launched (t = 21 s).

So, the value of $g_s$ is $g_s=\frac{30}{21} \:\frac{m}{s^2}$.