PLEASE HELP, this is a physics problem and i don't know how to solve it, it's due in 4 hours!!!?

Question

A stone is thrown straight up from the ground with an initial speed of 28 m/s. At the same instant, a stone is dropped from a height of h meters above ground level. The two stones strike the ground simultaneously. The acceleration of gravity is 9.8 m/s^2. Find the height h. Answer in units of m.

Answer 1

When stone #1 hits the ground, it will have the same speed it had when it left, 28 m/s, but negative compared to the direction it started.  So the total change in velocity is -56 m/(s^2).  If the stone is accelerating at -9.8 due to gravity, it will take 5.71 seconds to go up, and fall down.

So, how high up will a stone have to be that it take 5.71 seconds to fall?

.5at(^2) + h = 0

a = -9.8, t = 5.71, so solve for h.

This is the same answer as the person above, but the equation really is .5a(t^2) + v0t + h, not a(t^2) + v0t + x

Answer 2

The first matter of things to take care of is to determine how long the stone being tossed up takes to hit the ground again. The stone will go up (decreasing 9.8m/s^2) until it turns around and hits the ground- again at 28 meters per second. This equation, which you probably have is a linear function along the lines of:

v = at + v(initial)

The final v we're looking for is when its going 28m/s in the negative direction, and our initial speed is 28m/s in the positive direction. So moving that around and plugging in for acceleration you get:

-28 = -9.8t + 28

-56 = -9.8t

t= 5.714 seconds.

Now the other stone has to hit the ground in this same amount of time undergoing the same acceleration from high x. The equation for this (derived from calculus... you probably have it on some kind of equation sheet) is:

x = at^2 + v(initial)*t + x(initial)

Our initial velocity is zero since we're dropping it from rest, and the final x we're looking for is the ground which is also zero. So you just plug everything in and you're left with x(initial) being your height in meters.

0 = -9.8(5.714)^2 + 0 + x(initial)

x(initial) = 9.8(5.714)^2

x(initial) = solution (h) = 320 meters

Answer 3

First step is to determine the total flight time for the first stone (thrown vertically upwards). The time for the first stone to reach its peak is given by the formula

Vf - Vo = gT

where

Vf = final velocity = 0 (when stone reaches it peak)

Vo = initial velocity = 28 m/sec (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

T = time for stone to reach its maximum height

Substituting appropriate values,

0 - 28 = (-9.8)T

NOTE the negative sign attached to the acceleration due to gravity. It merely implies that the stone loses speed as it is going up.

Solving for T,

T = 2.86 sec.

Thus, the total flight time of the first stone = 2(2.86) = 5.72 sec.

This is also the flight time of the second stone (dropped from a height of "h" above ground level). To determine "h", the working formula is

h = VT + (1/2)(gT^2)

where

h = height at which the second stone was dropped

V = initial velocity of 2nd stone = 0 (since it was dropped)

T = 5.72 sec.

g = 9.8 m/sec^2

Substituting values,

h = 0(T) + (1/2)(9.8)(5.72)^2

h = 160.32 meters

Hope this helps.