A cannon is fired from a cliff 200 m high downward at an angle of 30o with respect to the horizontal. If the muzzle velocity is 34 m/s, what is its speed (in m/s) when it hits the ground?
To find the speed of the cannonball when it hits the ground, we can break down the motion into horizontal and vertical components.
First, let's find the time it takes for the cannonball to hit the ground. Since the cannonball is fired horizontally, there is no initial vertical velocity. Thus, we can use the following equation to find the time of flight:
h = (1/2) * g * t^2
where h is the initial vertical displacement (200 m) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Rearranging the formula to solve for time, we get:
t = sqrt(2h/g)
Substituting the given values, we have:
t = sqrt(2 * 200 / 9.8) = sqrt(40.816) = 6.39 seconds (approx.)
Now, to find the horizontal distance covered by the cannonball during this time, we can use the formula:
d = v * t
where d is the horizontal distance, v is the horizontal component of the velocity, and t is the time of flight.
Since the cannonball is fired at an angle of 30 degrees with respect to the horizontal, the horizontal component of the velocity (vx) can be found using:
vx = v * cos(theta)
where v is the muzzle velocity and theta is the angle of projection.
Substituting the given values:
vx = 34 * cos(30) = 34 * 0.866 = 29.524 m/s (approx.)
Now, substituting this value of vx and the time of flight (t) into the formula for horizontal distance (d), we get:
d = 29.524 * 6.39 = 188.615 m (approx.)
Finally, to find the speed of the cannonball when it hits the ground, we can use the Pythagorean theorem:
speed = sqrt(d^2 + h^2)
Substituting the values of d and h, we have:
speed = sqrt(188.615^2 + 200^2) = sqrt(35563.94 + 40000) = sqrt(75563.94) = 274.80 m/s (approx.)
Therefore, the speed of the cannonball when it hits the ground is approximately 274.80 m/s.