Asked by Anonymous
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 .
Answers
Answered by
Damon
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
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
Answered by
Henry
Anonymous, check your 2-2-11,12:26am post.
Answered by
Anonymous
I need to know the initial velocity and the units are meters
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