Answer:
Option"B" is correct.
Explanation:
when a body move with constant velocity then acceleration is zero.
a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how
long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175
s this is the time to fall from the top; it would take the same time to travel
upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175
= 0.35s
b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice
to solve this problem the time it takes to fall the final 0.13 m is: time it
takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to
fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it
takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m
is then twice this, or 0.08s
Answer:
Are you looking for the formula? because if so:
M L^2 T^-3 I^-2
Explanation:
May I have brainliest please? :)
Answer:
a) 2.7s
b) 29 m/s
Explanation:
The equation for the velocity and position of a free fall are the following
-(1)
- (2)
Since the hot-air ballon is <em>descending </em>at 2.1m/s and the camera is dropped at 42 m above the ground:
To calculate the time which it takes to reach the ground we use eq(2) with x=0, and look for the positive solution of t:
t = 2.71996
Rounding to two significant figures:
t = 2.7 s
Now we calculate the velocity the camera had just before it lands using eq(1) with t=2.7s
v = -28.782 m/s
Rounding to two significant figures:
v = -29 m/s
where the minus sign indicates the downwards direction
