Physics_Kinematics?

Question

Can anyone explain how to work out a problem like this?
1. A person throws a ball vertically downwards from the top of a building. the ball leaves the throwers hand with a speed 10 meters/second

A. Whats it the speed after 1 second? 2? 3 ?
B. How far does it fall in 1 second? 2? 3 ?
C. What is the magnitude of its veclocity after falling 10 meters ?

Answer 1

A)
What you know:
vi = 10 m/s
g = 9.8 m/s²
t = 1, 2, 3s
What you want to know: vf
Use one of the kinematic equations:
g = (vf - vi) / t ──► vf - vi = gt ──►vf = gt + vi
v(t) = 9.8*t + 10
v(1) = 19.8 m/s
v(2) = 29.6 m/s
v(3) = 39.4 m/s

B)
What you know: g, vi, vf, t
What you want: d
Again, use a kinematics equation (this time, you have 2 options; I'll just pick one):
g = (vf² - vi²) / (2d) ──► d = (vf² - vi²) / (2g)
d(1) = 14.9 m
d(2) = 39.6 m
d(3) = 74.1 m

C)
What you know: g, vi, d
What you want: vf
Use the kinematic equation I used in part (B):
g = (vf² - vi²) / (2d) ──► vf² = 2gd + vi²
──► vf = √(2gd + vi²)
vf = √(2*9.8*10 + 100) = 17.2 m/s

Answer 2

Velocity = accelleration x time
Velocity = speed and direction
magnitude = speed when travelling in one direction and disregarding negative direction (always positive answer)
average acceleration of gravity on Earth = 9.8 m/sec

10m/s + (9.8m/s squared x 1s) = 19.8 m/s
10m/s + (9.8m/s squared x 2s) = 29.6 m/s [etc]

Really, there's not enuf information to answer this since there is no information on the material of the ball - the above assumes the ball is in a void, ie, no air. In air, there would be a drag force that would lower the actual speeds. For small distances, this can be assumed to be zero. In the Gaileo experiment from the leaning tower, since both balls were of the same material, the air slowed them both down about equally.

Distance = velocity x time (D = v x t)

after 2s, 29.6m/s x 2s= 49.2 m [etc]

v = a x t = D/t or a x t = D/t
D = a x t x t or t = sqrt (10 s sq/9.8)
v = 10m/s + (9.8m/s sq x sqrt(10s sq/9.8)) which is close to 19.8 m/s

Answer 3

Gravitational Acceleration will be assumed to be: 9.8m/s^2

A) If the ball has an initial speed of 10m/s, after one second it will have accelerated another 9.8 m/s thus it's speed will be

10m/s + 9.8m/s = 19.8 m/s (roughly 20m/s)

B) The speed (velocity to be specific) is specified by:

V = G * T + Vi (where Vi is the initial velocity)

Thus at 1 second this equation computes:
V = 9.8m/s^2 * 1s + 10m/S = 19.8m/s

At 2 seconds it is:
V = 9.8m/s^2 * 2s + 10m/s = 29.6m/s

At 3 seconds it is:
V = 9.8m/s^2 * 3s + 10m/s = 39.4m/s

And so on and so on...

C) This one is a bit tricky. We have to use the equation that relates speed, acceleration, and time to distance. It is:

D = .5*G*T^2 + Vi*T

Using this equation, we can compute the time that will have passed after falling 10m.

10m = .5 * 9.8m/s^2 * T^2 + 10m/s * T

This can be reduced to a quadratic equation that looks like:

.5G*T^2 + Vi*T -10 = 0

or

4.9*T^2 + 10*T -10 = 0

Solve using the handy quadratic formula (look it up) and you get a time of approximately .735 seconds.

Plug that into our equation in Question B) and you get:

V = 9.8m/s^2 * .735s + 10m/s = 17.2 m/s

Not a simple solution, though.

Answer 4

With constant acceleration:
x(t)=x0 + v0*t + (1/2)*a*t^2
v(t)=v0+a*t
v(t)^2=v(0)^2+2*a*x

v=velocity (speed in meters per second)
x=displacement (distance in meters)
t=time (seconds)
v0=initial velocity
x0=initial displacement
a=acceleration

Let's say the top of the building is x=0, and x increases in a downward direction. The ball starts at v0 which is 10 m/s. We're going to ignore air friction and assume we're on earth where gravity is 9.8 m/s^2.

Part A...
v(t)=10+9.8*t
v(0)=10
v(1)=10+(9.8)(1)=19.8 m/s
v(2)=29.6 m/s
v(3)=39.4 m/s

Part B...
x(t)= 0 + 10*t + 4.9*t^2
x(0)=0
x(1)=0+10+4.9=14.9 m
x(2)=0+20+19.6=39.6 m
x(3)=0+30+44.1=74.1 m

Part C...
v^2=v0^2+2ax
v=sqrt(v0^2+2ax)
v=sqrt(100+196)
v=17.205 m/s (adjust as needed for significant figures)

Answer 5

Use the equations for projectile motion, but ignore the x components since everything is in the vertical plane.

Use:
V = Vo - gt for the velocity (part a)
y= yo + Vot -(1/2)gt^2 for position (part b)
You can actually set yo = 0. For part c just plug in 10 for Y and solve for
V^2 = Vo^2 - 2gy (part c) Plugging in 10 for y.

Answer 6

v = v0 +g*t
d = v0*t + 0.5*g*t^2