Answer:
0.215 [tex]\frac{m}{s^{2} }[/tex]
Explanation:
by using the equations of the constant acceleration motion or uniformly accelerated rectilinear motion:
V=[tex]v_0}[/tex] + a⋅t
X=[tex]X_0}[/tex]+[tex]v_0}[/tex] *t + [tex]\frac{a}{2} * t^{2}[/tex]
in this case [tex]X_0}[/tex] = 0, [tex]v_0}[/tex] = 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 - [tex]\frac{a}{2} * t^{2}[/tex]
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=[tex]\frac{2.5}{t}[/tex]
and replacing in 2. :
14.5=2.5 *t - [tex]\frac{2.5}{2t} *\ t^{2[/tex]
14.5=2.5 *t - [tex]\frac{2.5t}{2} [/tex]
14.5=[tex]\frac{2.5t}{2} [/tex]
by solving for t:
t = [tex]\frac{14.5 * 2}{2.5}[/tex]
t= 11.6 s
with this value of t,replacing in 3. :
a= [tex]\frac{2.5}{11.6}[/tex]
a= 0.215
