Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t = 21.2 s, total time of travel.
The vertical motion obeys the equation
s = ut - (1/2)gt²
When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²
Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.
Answer:
426.84 m
Explanation:
initial velocity u = 0
time t = 3.3 s
distance travelled s = 53.4 m
acceleration due to gravity = g
s = ut + 1/2 g t²
53.4 = 0 + 1/2 g x 3.3²
g = 9.8 m /s²
For the whole length of fall
distance travelled = h
total time = 6.6 + 3.3 = 9.9 s
h = ut + 1/2 g t²
u again = 0
h = .5 x 9.8 x 9.9²
= 480.24 m
distance travelled in last 6.6 s
= 480.24 - 53.4
= 426.84 m
It lets the viewer know it's something to do with underwater.
