Physics equation for this?

Question

A stone is thrown vertically upward with a speed of 10.0 m/s from the edge of a cliff 60.0 m high.

How much later does it reach the bottom of the cliff?

What is its speed just before hitting?

What total distance did it travel?

Answer 1

When the stone was thrown upwards,

Initial velocity = 10 m/s

Final velocity = 0 m/s

acceleration = -9.8 m/s/s

Distance travelled upwards can be solved by third equation of motion

v^2 - u^2 = 2as

s = (v^2 - u^2)/2a

  =  (0 - 100)/2(-9.8)

  = 5.1020408163265306122448979591837m

Time taken to reach that height can be calculated by using

v= u+at

t = (v-u)/a = (0-10)/(-9.8) = 1.0204081632653061224489795918367s

Height of the cliff = 60m

WHile falling down,

Total height from where it will fall = 60 + 5.1020408163265306122448979591837 = 65.1020408163265306122448979591837m

Now initial velocity = 0 m/s

using v^2 - u^2 = 2as

v^2 = 2as = 2 (9.8) (65.1020408163265306122448979591837)

here the acceleration will be 9.8 because the object is falling down.

     = 1275.9999999999999999999999999999

v = 35.721142198983503388631379410917 m/s

Using s = ut + 1/2 at^2 t can be solved

t = sqrt {2s/a}   as u =0

  = 3.6450145101003574886358550419303 s.

So total time it will take from the instant of throw to the instant of striking down is

         T = 3.6450145101003574886358550419303 + 3.6450145101003574886358550419303 = 7.29002902020071497727171008386 seconds

approximately 7.2 seconds.

Total distance travelled = 65.1020408163265306122448979591837 + 5.1020408163265306122448979591837 = 70.204081632653061224489795918363 m

approximately 70. 204 metres.

am I right?

Answer 2

0=v-g*t=10-9.8*t

t=10-9.8*t

solving for t:

10 / 9.8 = 1.02 seconds until zero velocity "max height"

s= v*t-1/2gt^2

s=(10*1.02)-0.5*9.8*(1.02)^2

= 5.1 meter "from top of cliff to max height"

maximum height from ground=60+5.1 meter

so it falls 65.1 meter

time from max height to ground:

s=1/2*g*t^2

t=sqr(2*s/g)

=sqr((2 * 65.2) / 9.8) = 3.65 seconds

total round trip time:

"how much later from start of throw"

3.65+1.02=4.67 seconds

from conservation of energy

"speed just before hitting"

mgh=1/2*m*v^2

v=sqr(2*9.8*65.1)

=sqr(2 * 9.8 * 65.1) = 35.72 m/s

total distance traveled= 60+5.1+5.1

60 + 5.1 + 5.1 = 70.2 meters

Answer 3

Basic kinematics equations

Acceleration due to gravity = -9.8m/s^2

initial velocity = 10m/s

To do this problem, I would break it into two parts

The first part would be calculating when it's velocity became 0 (in order to begin falling, this would have to occur at some point.)

To do that:

vf = vi + at

where vf = 0

vi = 10m/s

a=-9.8m/s^2

solve for t

0 = 10 - 9.8t

-10 = -9.8t

divide both sides by -9.8

which ~ 1.02s

Now get the distance it travels in that time using

d = vit + (.5)at^2

d = 10(1.02) + (.5)(-9.8)(1.02)^2

d = 10.2 - 4.9(1.0404)

d ~ 5.1m

That gives you total distance

5.1x2(up and down) + 60 =70.2

To get the time it takes to go from 65.1 to the bottom do

d = vit + (.5)at^2

(distance is negative since we're going down)

-65.1 = 0t + .5(-9.8)t^2

-65.1 = -4.9t^2

13.3 = t^2

t ~ 3.64s

To get speed just before hitting (vf)

vf^2 = vi^2 + 2ad

(remember, speed is a scalar, so there's no direction, therefore no negative)

vf^2 = 0^2 + 2(9.8)(65.1)

vf^2 = 0 + 1275.96

vf ~ 35.7m/s