Question

A human expedition lands on an alien planet. One of the explorers is able to jump a maximum distance of 14.5 m with an initial speed of 2.50 m/s. Find the gravitational acceleration on the surface of the alien planet. Assume the planet has a negligible atmosphere. (Enter the magnitude in m/s2.)

Answer 1

Answer:

0.215 $\frac{m}{s^{2} }$

Explanation:

by using the  equations of the constant acceleration motion or uniformly accelerated rectilinear motion:

V=$v_0}$ + a⋅t  

X=$X_0}$+$v_0}$ *t + $\frac{a}{2} * t^{2}$

in this case $X_0}$ = 0, $v_0}$ = 2.5 and X= 14.5.

replacing those numbers in the equations,and knowing that a gravitational acceleration is always negative :

1.   V= 2.5 - a⋅t

2.   14.5=2.5 *t - $\frac{a}{2} * t^{2}$

we have 2 unknowns (g and t) ,so this can be solve by the Substitution Method :

0= 2.5 - a⋅t  

V in this case will be 0,because the speed at the maximum distance is 0.

by solving for a:

3.   a=$\frac{2.5}{t}$

and replacing in 2. :

14.5=2.5 *t - $\frac{2.5}{2t} *\ t^{2$

14.5=2.5 *t - $\frac{2.5t}{2}$

14.5=$\frac{2.5t}{2}$

by solving for t:

t =  $\frac{14.5 * 2}{2.5}$

t= 11.6 s

with this value of t,replacing in 3.  :

a= $\frac{2.5}{11.6}$

a= 0.215