Question
A cannon fires a cannot into the ocean off a cliff. The cannon is angled at 0 degrees to the horizontal.
The cannonball is fired at an initial velocity of 120 m/s and the cliff is 35 m above the water level. If the
water is considered to be where the height = 0, which of the following kinematic equations would be
used to find the time the cannonball? (1 point)
answers:
V^2 = v0^2 +2*a*x
x= v0*t + 1/2*a*t^2
x=((vO+v)/2)*t
v = x/t
The cannonball is fired at an initial velocity of 120 m/s and the cliff is 35 m above the water level. If the
water is considered to be where the height = 0, which of the following kinematic equations would be
used to find the time the cannonball? (1 point)
answers:
V^2 = v0^2 +2*a*x
x= v0*t + 1/2*a*t^2
x=((vO+v)/2)*t
v = x/t
Answers
GPT-4o mini
To find the time it takes for the cannonball to reach the water level from the top of the cliff, we can use the kinematic equation that relates displacement, initial velocity, acceleration, and time.
Given that the cannonball is fired horizontally, the relevant kinematic equation to use is:
**x = v0 * t + 1/2 * a * t²**
In this case:
- \(x\) is the vertical displacement (which will be -35 m since it's falling down),
- \(v_0\) is the initial vertical velocity (which is 0 m/s since it's fired horizontally),
- \(a\) is the acceleration due to gravity (approximately -9.81 m/s², negative because it is directed downward),
- \(t\) is the time.
Thus, the correct answer is:
**x = v0 * t + 1/2*a*t²**
Given that the cannonball is fired horizontally, the relevant kinematic equation to use is:
**x = v0 * t + 1/2 * a * t²**
In this case:
- \(x\) is the vertical displacement (which will be -35 m since it's falling down),
- \(v_0\) is the initial vertical velocity (which is 0 m/s since it's fired horizontally),
- \(a\) is the acceleration due to gravity (approximately -9.81 m/s², negative because it is directed downward),
- \(t\) is the time.
Thus, the correct answer is:
**x = v0 * t + 1/2*a*t²**
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