A cannon ball is fired horizontally with a velocity of 50m/s from the top of a cliff of 90mh. after how many seconds will it strike the plane at the foot of the cliff and at what distance at the foot of the cliff will it strike?
time to hit:
4.9t^2 = 90
t = 4.29 s
4.29s * 50 m/s = 214.28 m
To find the time it takes for the cannonball to strike the plane at the foot of the cliff, we can use the equation of motion for vertical motion:
y = yo + vyo*t - 0.5*g*t^2
Where:
- y is the vertical displacement (distance) from the top of the cliff,
- yo is the initial vertical displacement (height of the cliff),
- vyo is the initial vertical velocity (0 m/s, since the cannonball is fired horizontally),
- t is the time, and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since the cannonball is fired horizontally, the initial vertical velocity (vyo) is zero. Therefore, the equation becomes:
y = yo - 0.5*g*t^2
We can substitute the given values into the equation:
90m = 0 - 0.5 * 9.8m/s^2 * t^2
Simplifying the equation, we get:
45m = 4.9m/s^2 * t^2
Now, let's solve for t by rearranging the equation:
t^2 = (45m) / (4.9m/s^2)
t^2 ≈ 9.18
Taking the square root of both sides, we get:
t ≈ √9.18
t ≈ 3.03 seconds
So, it will take approximately 3.03 seconds for the cannonball to strike the plane at the foot of the cliff.
Now, let's find the horizontal distance at the foot of the cliff where the cannonball will strike. Since the cannonball is fired horizontally, we can use the formula:
distance = velocity * time
Given that the velocity is 50 m/s and the time is 3.03 seconds, we can calculate the distance:
distance = 50 m/s * 3.03 s
distance ≈ 151.5 meters
Therefore, the cannonball will strike the plane at a distance of approximately 151.5 meters at the foot of the cliff.