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

0.215
`(m)/(s^(2) )`

Step-by-step 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 +
`(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 -
`(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=
`(2.5)/(t)`

and replacing in 2. :

14.5=2.5 *t -
`(2.5)/(2t) *\ t^{2`

14.5=2.5 *t -
`(2.5t)/(2)`

14.5=
`(2.5t)/(2)`

by solving for t:

t =
`(14.5 * 2)/(2.5)`

t= 11.6 s

with this value of t,replacing in 3. :

a=
`(2.5)/(11.6)`

a= 0.215