Physics problem.. please answer...?

Question

a man dropped a coin on top of a 150m. building.

a) find the total time of flight

b) how far will it travel after 3 sec?. 5sec.? 7 seconds?

c) Find the time needed to take half of its flight?

d) what is the acceleration due to gravity?

Answer 1

first of all u correct ur question!

what do u mean by, on the top of the building???

n also u did not give the weight of the coin.

Answer 2

gravity is 9.8 metres per second squared and every second it increases by 9.8 metres so you can work it out from there.

Answer 3

is been a while since i have done high school physics but if i remember correctly this is it. but i have made a few assumptions as you have not been heaps clear with the question. i have mentioned the assumptions

at the earths surface gravity is 9.8ms^-2

we use the formula s = Vi + 1/2at^2     (Vi = initial velocity, a= acceleration, t=time, s= distance)

so if we assume the coin was originally at rest. (ie. dropped and not thrown) then;

a) 5.53s

b)it will travel 44.1m. 122.5m. 240.1m

c)3.91 secs

d)9.8ms-2

all these answers were obtained using the formula stated above and subbing in the numbers.

Answer 4

a) 8 secs

b)it will travel 80m. 110m. 148

c)4 secs

d)120mph

Answer 5

if the building is on mars:

a) 3 hours + give or take two tequilla shots

b) 12 sucks (the metric system on mars; 1 suck = 12.3256 ft.) /  22 scks / 37 scks

c) almost 17 days, as NASA will be mostly interested in the coin, and will send a shuttle do make some tests, which will considerably slow the fall

d) acceleration depends on how drunk the marsian was that dropped it from the building instead of trying to pay with it a heineken beer

hoping for best answer, of course ;)

Answer 6

I can't answer u'r first Q as u've not mentioned the value of U initial velocity...so the next 3 r dependant on that..

for the last Q it's 9.8 m/s square....anywhere on the surface of the earth!

Answer 7

a) s = ut +½*a*t²

150= 0+½ *9.81*t²

t² = 150/4.905

t² = 30.58

t = 5.53 seconds

b)s = ut +½*9.81*3²

s = 0+ 4.905*9

s = 44.15meters

You can do the distance for time 5 seconds . Note that the coin will have fallen the full 150m and hit the ground before 7 seconds.

c) s= ut + ½*at²

75 = 0 + 4.905t²

t² = 75/4.905

t² = 15.3

t = 3.91 seconds

d) Acceleration = 9.81m/s² downwards.

Answer 8

Δx = 150 m

a) find the total time of flight

Δx = vo Δt + ½ a Δt²

150 = 0 + 1/2 * g * Δt²

150 = 1/2 * 9.8 * Δt²

150 / 4.9 = Δt²

Δt = √ 30.61 = 5.53 s

b) how far will it travel after 3 sec?. 5sec.? 7 seconds?

Δx = vo Δt + ½ a Δt²

3 sec; Δx = 0 + ½  * 9.8 * 3² =  44.1 m

5 sec; Δx = 0 + ½ * 9.8 * 5² = 122.5 m

7 seconds; = 150 m ; the coin is on the ground after 5.53 secs

c) Find the time needed to take half of its flight?

Δx / 2 = vo Δt + ½ a Δt²

150 / 2 = 0 + 1/2 * 9.8 * Δt²

75 / 4.9 = Δt²

Δt = √ 15.3 = 3.91 s

d) what is the acceleration due to gravity?

9.8 m/s/s

All of this neglects the effect of air resistance and assumes the coin has no initial velocity. (coin dropped, not thrown down)