A ball is dropped from 80 meters how high from the ground is it at 1,2,3, and 4 seconds?

Answer 1

Let h = height from the ground at time t (in second).
dy = distance fallen at time t (in second).

Therefore,
h = 180 - dy, where dy = (1/2)gt^2

a) at t = 1 sec; h = 80 - (1/2)(9.8)(1)^2 = 75.1 m ANS

b) at t = 2 sec; h = 80 - (1/2)(9.8)(2)^2 = 60.4 m ANS

c) at t = 3 sec; h = 80 - (1/2)(9.8)(3)^2 = 35.9 m ANS

d) at t = 4 sec; h = 80 - (1/2)(9.8)(4)^2 = 1.6 m ANS

Hope I help.

teddy boy

Answer 2

Distance fallen after t seconds = .5 * g * t ** 2
Assume g = 10 ms-2.
Then distance fallen = 5 * t ** 2 and height = 80 - 5 * t ** 2
so evaluate this when t=1, t=2, t=3 and t=4;
after 1 second, height = 80 - 5 = 75m.;
after 2", height = 80 - 20 = 60m.;
after 3", height = 80 - 45 = 35m.;
after 4", height = 80 - 80 = 0m. (this is when it hits the ground.)

Answer 3

The distance it falls each second is greater because of the acceleration of gravity, according to the formula d = 1/2 g t^2.

Plugging in the numbers, after one second it has fallen 4.9 meters; 4.9*4 or 19.6 meters after two seconds; 4.9*9 or 44.1 meters after three; and 4.9*16 or 78.4 meters after four seconds.

If your instructor told you to use 10 m/s/s for g, these answers would be 5, 20, 45, and 80 - or 75, 60, 35 and zero if you want the height above the ground.

Answer 4

Distance is 80 - (g t²)/2 assuming negligible air resistance

g ~ 9.8 m/s²

for t = 1, 2, 3, 4s
distance = 75.1, 60.4, 35.9, 1.6m

Answer 5

the rate of acceleration is 32 feet per second squared... so it will hit before some of your stated time limits, will it not?

32 ft, 128 ft, 288 ft, 512 ft..

that works out to 160 feet the second second, and 440 feet the third, but you've already hit the ground... so to answer your question... you can't get there from here!

Answer 6

I don't have the time to figure it out, but here is the formula for someone who does. The ball will accelerate at the rate of 32 feet per second per second.