Physics help, please? A rock is falling.?

Question

A rock dropped from 1200 ft., ignoring air resistance. How long does it take to hit ground? How fast will it be traveling at impact? Would these answers change if it were thrown 90 mph horizontally? How far would the horizontal rock go before impact?

I have the 32 ft/sec^2. I just don't seem to be doing something correct. I think it hits at the same time. Then I just convert the 90 mph into sec for the first answer.

Answer 1

The first part can be answered from:

s=ut+0.5at^2

Where s is the final velocity, u is the initial velocity, a is acceleration due to gravity and t is time. As the rock has been released from rest, then u =0, so that

s= 0.5at^2

This could be rearranged to yield :

t- sqrt (2*s/a)= sqrt (2*1200/32)=8.66 seconds.

The velocity at impact can be calculated from:

v^2 = u^2+2as, where v is the final velocity.

v= sqrt (2*a*s) , as u is zero.

v = sqrt (2*32*1200)= 277.128ft/s

The rock would go 132 ft/s (90mph) at at time of 8.66s, giving 1143 ft.

Answer 2

I agree, tell your instructor to use metric. The more that people become familiar with it, the sooner we can throw out the stupid imperial system.

There are a few ways to go at this problem, This is the one that I think is the slickest. First, to avoid confusion you should convert units. 1200ft = 365.76m, g = 9.81 m/s/s, 90 mi/hr = 40.23 m/s

I'll answer the second question first. According to conservation of energy, the rock's final kinetic energy (Ek) should equal the gravitational potential energy (U) before it was dropped. given:
m = mass
h = original height
g = acceleration due to gravity;

U = m*g*h = -Ek = (-1/2)*m*v^2. so solving for "v" gives:

v = -sqrt(2*g*h)! So mass is not really necessary to know.

Now that we know the final speed, we can switch to using momentum (P). In the case of a CONSTANT force F = mg, where t = time:
P = m*v = F*t = m*g*t,
so again dropping mass and solving using our previous velocity equation,

t = v/g = -sqrt(2*g*h)/g !

The third question is easy. According to the Superposition principle; for a free body, forces and velocities in the vertical have no affect on the forces and velocities in the horizontal(i.e at 90 degrees to each other).
So in order to find the horizontal distance, just multiply the horizontal speed by the total time t.

Finally reconvert units to imperial(if you like)

Well I hope that helps you, Good Luck!
~Donkey Hotei

Answer 3

Use the equation for freely falling bodies:

h = vi t + 1/2 g t^2

vi = 0 since the initial vertical velocity is zero. So you can solve for t.

For the next question use:

vf^2 - vi^2 = 2 g h

vi is still zero.

Solve for vf.

If it were thrown at 90 mph horizontally, the falling time will not change since the initial velocity has no vertical component, so the time for its vertical descent will not be affected. But the velocity at impact will change.

The velocity at impact for these will be: (Pythagorean theorem)

v = sq rt (vf^2 + v horizontal^2)

The horizontal velocity does not change since acceleration is in the vertical direction. It is still 90 mph at impact.

Convert mph to ft/s first.

Answer 4

Time to ground use d=1/2 a t^2 (distance= 1/2 acceleration x time squared)


speed at impact use speed= acceleration x time and use the time you from before


those won't change if a horizontal speed is added


distance = rate x time for the last part, just be sure to put 90 mph into miles per second




p.s. tell your instructor to use metric

Answer 5

s = ut + ft^2/2
1200 = 0 + 16t^2
so t = 8.66secs

v^2-u^2 = 2fs
v^2 - 0 = 2*32*1200 = 76800
so v = 277 ft/sec

Yeh, 90 mph = 132 ft/sec, so it will travel horizontally 8.66*132 = 1143 feet.