A cannonball is fired horizontally from the top of a cliff. The cannon is at height = 70.0 above ground level, and the ball is fired with initial horizontal speed . Assume acceleration due to gravity to be = 9.80 .

Anonymous, check your 2-2-11,12:26am post.

I need to know the initial velocity and the units are meters

Well, well, well. We've got a cannonball flying horizontally off a cliff! That's definitely an interesting situation. Let's see what we can figure out here.

First off, we've got a height of 70.0 above ground level. That's pretty high up, so I hope our cannonball isn't afraid of heights!

Now, we need to know the initial horizontal speed. Unfortunately, you forgot to tell me what it is. Maybe the cannonball is feeling shy and doesn't want to reveal its speed?

Finally, we've got the acceleration due to gravity, which is a pesky old 9.80. Gravity just loves to mess things up, doesn't it?

So, while I can't give you a specific answer without knowing the initial horizontal speed of the cannonball, I can definitely help you figure out some other things. Just let me know what you want to know, and I'll do my best to make you laugh while giving you the information!

To find the time it takes for the cannonball to hit the ground, we can use the kinematic equation for vertical motion:

h = h0 + v0t - (1/2)gt^2

Where:
h = final height (in this case, 0 because the ground is at height 0)
h0 = initial height (given as 70.0m)
v0 = initial vertical velocity (unknown)
g = acceleration due to gravity (-9.80 m/s^2)
t = time (unknown)

Since the cannonball is fired horizontally, its initial vertical velocity (v0) is zero.

Using the equation above, we can rearrange it to solve for time:

0 = 70.0 + 0t - (1/2)(-9.80)t^2

Simplifying:

0 = 70.0 - 4.90t^2

Rearranging again:

4.90t^2 = 70.0

Dividing both sides by 4.90:

t^2 = 14.29

Taking the square root of both sides:

t = √14.29

t ≈ 3.78 seconds

Therefore, it takes approximately 3.78 seconds for the cannonball to hit the ground.

Well, you did not say what the question is but you have two problems, a vertical problem and a horizontal problem

The vertical problem is the ball falls 70 meters (I assume) in t seconds
The horizontal problem is he ball proceeds at constant initial speed u for the same t seconds.
Vertical
initial vertical speed = 0
initial height = 70
final height = 0
acceleration = -g = -9.8 m^s^2
0 = 70 + 0 t + (1/2)(9.8) t^2
4.9 t^2 = 70
t = 3.78 seconds in the air

Now you did not say what initial speed u was, although I am sure the problem in the book did so and also included the missing units. So
distance = rate * time = u * 3.78 meters