Physics Question!? PLZ HELP!?

Question

1. A stone is thrown horizontally at a speed of 5m/s from the top of a cliff 78.4m high.

a) How long does it take the stone to rech the bottom of the cliff?

b)How far from the base of the cliff does the stone strike the ground?

c) What are the horizontal and vertical components of the velocity of the stone as it hits the ground?

d) What is the final speed of the stone as it hits the ground?
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2. How would the answers to a,b,and c change:

a) If the stone were thrown with twice the horizontal speed? (Explain)

b) If the stone were thrown with the same speed but the cliff were twice as high? (Explain)

The answers to 1 are:
a) 4.0 s

b) 20 m

c)horizontal - 5 m/s
vertical - 39.2 m/s

d) 39.5 m/s

I know these answers, but do you do them?

Answer 1

1.
a. s = (1/2)gt^2 where g = acceleration of gravity. the horizontal speed does not affect how long it will take to reach the ground.

s = (1/2)(9.8)t^2 = 78.4
t^2 = 16 and t = 4 seconds

b. Now use the horizontal speed. The distance will be this speed times the time from 1.

distance = 5*4 = 20 meters

c. Horizontal is 5 meters per second.
Vertical is caused by the acceleration of gravity
v = gt = 9.8*4= 39.2

d. speed = SQRT(39.2^2 + 5^2)
speed = 39.52 m/s

2.
a. Time to reach the ground would remain the same since this is in the vertical direction and a change in speed in the horizontal will have no effect.

It will land twice as far from the cliff since the time is still the same but the horizontal speed has doubled.

The vertical will be the same as mentioned above but the horizontal will double to 10. The horizontal speed is not affected by gravity and it has no acceleration, it is constant.

b. All would change except the horizontal speed which is a constant. Changing the height will change the time it takes the stone to reach the ground which in turn changes it's velocity at the time it reaches the ground. The horizontal distance will also change since it depends on time. The time to reach the ground depends on the square root of the height so it will go up by a factor of SQRT(2). vertical velocity at impact will also go up by this factor since it depends directly on the time as will the horizontal distance since it to depends directly on the time.

Answer 2

1.)consider,
h=78.4m
u(h)=5m/s

a.) now the time taken,t = sqrt(2h/g)
= sqrt(2*78.4/9.8)
= sqrt(16)= 4s

b.) the distance from the cliff, range R= u(h)*t
=5*4=20m

c.) the horizontal velocity is the same,it does not change.
therefore, u(h)= 5 m/s
u(v)= gt = 9.8*4=39.2 m/s

d.) the resultant velocity, v =sqrt{(gt)^2}
=sqrt{(9.8*4)^2}

2.) a.)substituting the vaues in formulae above,
time wdnt change.
range is doubled.
vertical velocity also doesnt change.

b.)time wd b increased root of 2 times.
range n vertical velocity doesnt change.

Answer 3

The equations of motion as derived from newtons second law are: X(t)=.5at^2+vt+x; Y(t)=.5at^2+vt+y; and V(t)=at+v
1.

a) 0=Y(t)=-.5gt^2+78.4
solve for t

b)plug your answer from part a into X(t)=5t and solve for X(t)

c) Vx=78.4; Vy=-.5gt plug answer from part a into t and solve for Vy

d) Vx, Vy, and speed form a right triangle. pathagorean theorm : speed =sqrt(Vx^2+Vy^2)

2.

A)look at the equations for an explanation
a) same
b)twice as far
c)Vy same, Vx doubled
d)use pathagorean theorm again

B) I'll let you do this one on your own. You should have no trouble if you work through the equations above...

Good Luck

Answer 4

Make a Chart.
.............X component..../....Y Component
Vi................5m/s............/.....0m/s
Vf................5m/s............/.....?
a ................0m/s^2......../.....9.8m/s^2
Di.................0m............./......0m
Df .................? ............../......78.4m
t .................................?.......

Acceleration in Y is 9.8m/s^2
Initial Distance in Y is 0m
Initial Velocity in Y is 0m/s^2

Make the origin at the point at which the object is released, and make the x and y directions in which it moves positive.

Use this eqn to answer part a:
change in D= 0.5at^2+Vit
78.4=0.5(9.8)t^2
t^2=16
so t=4

Now that you can fill in that part of the chart you can use the eqn (change in D=0.5at^2+Vit) in reference to the information in the X-component column to find answer b.

For part c: You already know the x-component's final velocity.
Use the eqn Vf=Vi+at in respect to the y-component information to find the y-component's final velocity.

For part d: Make a right triangle with the X and Y component velocities as legs and find the measurement of the hypoteneuse.