Answer: 1.77 s
Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:
xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then
(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s
then by solving the quadratric equation in t
t=1.77 s
1.A) 4.9 m
AL2006 Ace
The instant it was dropped, the ball had zero speed.
After falling for 1 second, its speed was 9.8 m/s straight down (gravity).
Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.
Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.
ANYTHING you drop does that, if air resistance doesn't hold it back.
Read more on Brainly.com - brainly.com/question/11776597#readmore
2 idk sorry
Answer:
I guess the acceleration would be 8 meters a second
Explanation:
I can't think of any other fitting way to put the answer sorry if it's not right
Answer: E = 941738.537J
Explanation:
to begin,
given that the mass = 2320 pound = 1052.334 kg
Δh = 110 ft = 33.528 m
given that distance (d) = 1283 ft = 391.058 m
also the speed (v) is 65 mph = 29.058 m/s
force (F) = 87 pounds = 386.995 N
from our knowledge in work energy theory;
E = Fd + 1/2mv² + mgh
E = (386.995 × 391.058) + (1/2×1052.334×29.058²) + (1052.334×9.81×33.528)
E = 151337.491 + 444278.2 + 346122.84
E = 941738.537J
i hope this helps, cheers.
