A cannon ball is fired horizontally with velocity of 15m from the top of a cliff 90m high.after how many seconds will it hit the foot of the cliff(a)time (b)at what distance from the foot of the cliff will it strike.
Why did the cannon ball audition for the role of a stuntman? Because it wanted to make a cliff-hanging entrance!
Now, to answer your question, let's use a bit of physics. When the cannonball is fired horizontally, its vertical velocity is zero. Therefore, the only force acting on it is gravity, which causes it to drop vertically.
We can use the equation for vertical displacement to find the time it takes for the cannonball to hit the foot of the cliff.
The equation for vertical displacement is:
h = (1/2)gt^2
Where:
h is the vertical displacement (90m),
g is the acceleration due to gravity (approximately 9.8 m/s^2),
and t is the time it takes for the cannonball to hit the foot of the cliff.
Plugging in the values:
90 = (1/2)(9.8)(t^2)
Simplifying the equation:
90 = 4.9t^2
Dividing both sides by 4.9:
t^2 = 18.37
Taking the square root of both sides:
t ≈ 4.28 seconds
So, the answer to part (a) is approximately 4.28 seconds.
Now, for part (b), we can calculate the horizontal distance the cannonball will travel using its initial horizontal velocity of 15 m/s and the time it takes to hit the foot of the cliff, which we just found to be approximately 4.28 seconds.
The equation for horizontal distance traveled is:
d = v * t
Where:
d is the horizontal distance traveled,
v is the initial horizontal velocity (15 m/s),
and t is the time it takes for the cannonball to hit the foot of the cliff (approximately 4.28 seconds).
Plugging in the values:
d = 15 * 4.28
Calculating the result:
d ≈ 64.2 meters
So, the answer to part (b) is approximately 64.2 meters from the foot of the cliff.
To find the time it takes for the cannonball to hit the foot of the cliff and the distance it will strike from the foot of the cliff, we can use the equations of motion.
First, let's analyze the horizontal motion of the cannonball. Since it is fired horizontally, there is no initial vertical velocity. Therefore, the only force acting on the cannonball in the horizontal direction is gravity, which does not affect its horizontal motion. Hence, the horizontal velocity remains constant throughout the motion.
The horizontal velocity of the cannonball is given as 15 m/s. Let's denote this velocity as V_x.
(a) Finding the time it takes for the cannonball to hit the foot of the cliff:
Since the horizontal velocity remains constant, we can use the equation:
distance = velocity × time
The distance covered horizontally is the same as the distance from the top of the cliff to the foot of the cliff, which is the horizontal range. Thus, we have:
range = V_x × time
Substituting the given values, where range = 90 m and V_x = 15 m/s, we can solve for time:
90 m = 15 m/s × time
To find the time (t), we can rearrange the equation:
time = range / V_x
Plugging in the values, we have:
time = 90 m / 15 m/s = 6 s
Therefore, it will take 6 seconds for the cannonball to hit the foot of the cliff.
(b) Finding the distance from the foot of the cliff where the cannonball will strike:
As mentioned earlier, the horizontal velocity remains constant. So, the cannonball will continue to move with a constant velocity of 15 m/s.
Using the equation for distance covered horizontally:
distance = velocity × time
Substituting the given values, where velocity (V_x) = 15 m/s and time = 6 s, we can find the distance covered by multiplying the velocity and time:
distance = 15 m/s × 6 s = 90 m
Therefore, the cannonball will strike the ground at a distance of 90 meters from the foot of the cliff.
Xo = 15 m/s.
h = 90 m.
g = 9.8 m/s^2
a. h = 0.5g*t^2 = 90 m. Solve for t.
b. Dx = Xo * t