Answer:
10.38m
Explanation:
Using the equation of motion formula
∆S = ut+1/2at²
S2-S1 = ut+1/2gt²
u is the initial velocity
t is the time taken = 2secs
g is the acceleration due to gravity = 9.81m/s²
30-S1= 0(2)+1/2(9.81)2²
30- S1 = 0+4.905(4)
30 -S1 = 19.62
-S1 = 19.62-30
-S1 = -10.38
S1 = 10.38m
The distance closest to how far it should have fallen after 2secs is 10.38m
