Someone help me with these math problems. Show me how to set up the equations and work them out.?

Question

1- a cannon launches a cannonball horiziontally off a cliff that is 25.4 meters high. If the cannonball is fired with a velocity of 28.3 meters per second, how far from the edge of the cilff will the ball land?
2- a stone is dropped from a cliff to the stream below. It takes 6.28 seconds for the stone to hit the water. A- How fast in meters per seconds is the stone traveling when it hits the water? B- How high in meters if the cliff above the stream?
3- a penny is dropped from a 700 ft. high bridge. A-How long does it take for the penny to reach the ground below? B- What is the velocity of the penny upon impact?
4- it takes 4.20 seconds for a ball that was thrown down from the top of a building to reach the sidewalk below. The balls velocity upon striking the sidewalk is 148 miles per hour. A- How tall is the building? B- what is the initial velocity of the ball?

Answer 1

There are essentially three equations that are used to solve all of these types of problems. These are called kinematics equations as they deal with moving objects experiencing a constant acceleration. It is important to keep track of horizontal and vertical movement when using these equations. These equations are:

2 a x = v^2 - v0^2

v = v0 + a t

x = x0 + v0 t + 1/2 a t^2

where x = position
x0 = initial position
v = velocity
v0 = initial velocity
a = acceleration
t = time

You simply read the problem, find out what you know, figure out what you want to find, select the appropriate equation, and do the math.

I will use your first one as an example. (Note: I will use vx or vy to differentiate between vertical (y) and horizontal (x) motion).

I know the following:

vx0 = 28.3 m/s vy0 = 0 m/s
x0 = 0 m y0 = 25.4 m
ax = 0 m/s^2 ay= -9.8 m/s^2
x =? y = 0 m
tx = ? ty = ?

I want to find x, but will need to know t to do so. The time it takes for the fall is equal to the time it travels horizontally.

tx = ty

Once I find this time I can find x. I need to use the horizontal information to find time and will select the following equation to help me.

y = y0 + v0y t + 1/2 ay t^2, y and v0y both are zero so this equation becomes

0 = y0 + 1/2 a t ^2, which can be solved for t.

t = (-2 y0 / a)^(1/2) = (-2 * 25.4 m / -9.8 m/s^2)^(1/2) = 2.28 s

I can use this new t and the information about the horizontal motion with the same equation to find x.

x = x0 + v0x t +1/2 ax t^2 where ax and x0 both equal zero

x = v0x t = 28.3 m/s * 2.28 s = 64.4 m

The ball travel 64.4 m in problem 1.

Answer 2

1) time = sqrt(2*25.4/9.8) = sqrt(5.183) seconds = 2.3 s

28.3 m/s * 2.3 s = 65 meters (ignoring wind resistance)

2) 6.28 s * 9.8 m s/s = 61.5 m/s

6.28 s = sqrt(2 * h / 9.8)
39.4384 = 2 * h / 9.8
386.5 = 2h
h = 193.25 meters

3) t = sqrt(2*700/32.2) = 43.5 seconds

43.5 * 32.2 ft s/s = 1400.7 ft/s (ignoring wind resistance)

you should be able to do #4 now