A cannon is fired from a cliff 190 m high downward at an angle of 21° 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?
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To find the speed of the cannon when it hits the ground, we need to break down the initial velocity into horizontal and vertical components.
The horizontal component of velocity remains constant throughout the motion because there is no horizontal acceleration. We can find the horizontal component of velocity using trigonometry:
Horizontal component of velocity (Vx) = Initial velocity (V) * cos(angle)
Vx = 34 m/s * cos(21°)
Next, we need to find the vertical component of velocity. Gravity acts in the downward direction, so the cannonball experiences a constant acceleration of -9.8 m/s^2. We can find the vertical component of velocity using trigonometry:
Vertical component of velocity (Vy) = Initial velocity (V) * sin(angle)
Vy = 34 m/s * sin(21°)
Now, we have the initial vertical velocity and the acceleration due to gravity. We can use the following kinematic equation to find the time it takes for the cannonball to hit the ground:
Vy = Initial velocity + (acceleration * time)
Since the cannonball is launched downward, the initial velocity is negative, so we can rewrite this equation as:
-34 m/s * sin(21°) = 0 + (-9.8 m/s^2) * time
Solving for time:
time = (-34 m/s * sin(21°)) / (-9.8 m/s^2)
Once we have the time it takes for the cannonball to hit the ground, we can calculate the distance traveled horizontally using the formula:
Distance = Horizontal component of velocity * time
Distance = Vx * time
Finally, we can find the speed when the cannonball hits the ground by calculating its resultant velocity using the Pythagorean theorem:
Speed = sqrt((Horizontal component of velocity)^2 + (Vertical component of velocity)^2)
Speed = sqrt((Vx)^2 + (Vy)^2).
Plugging in the values, we can calculate the speed when the cannonball hits the ground.