Answer:
Explanation:
Height reached = 164m
If velocity of launch be u
u² = 2gh
= 2 x 9.8 x 164
= 3214.4
u = 56.7 m /s
difference in u = 1/2 (diff in h / h) x u
1/2 x 1/164 x 56.7
.17 or .2
u = 56.7 ± .2 m /s
or 57± 1
b ) During the displacement of .300 m , there is acceleration to achieve velocity of 56.7 m /s
v² = 2as
a = v² / 2s
= 56.7² / (2 x .3)
= 5358.15 m /s²
diff in a
= 1/2 x (.02 / .3) x 5358.15
178.6
a = 5358.1 ±178.6
c ) force on the ground
mass ( g + a )
= 100 x ( 9.8 +5358.1)
= 536790 N
error in force = (.1/100 + 178.6 / 5358 ) x 536790
=( .01 +.03)x 536790
= 21471
(536790 ± 21471) N