A cannon is fired from a cliff 190 m high downward at an angle of 38o with respect to the horizontal. If the muzzle velocity is 41 m/s, what is its speed (in m/s) when it hits the ground?
Which equations do I need for this?
V^2 = sqrt[Vx^2 + Vy^2]
Vx remains 41 m/s * cos 38 during flight.
Vy at impact can be calculated using
Vy^2 = (V sin 38)^2 + 2 g H
V is 41 m/s, right?
I keep getting 64.28 m/s as my answer, but it's not right. What am I doing wrong?
To solve this problem, you can use the equations of projectile motion. In particular, we can use the equations for horizontal and vertical components of motion separately.
Let's break down the motion of the cannonball into horizontal and vertical components:
1. Horizontal motion:
The horizontal component of velocity remains constant throughout the motion since there is no horizontal acceleration. Therefore, the horizontal velocity, Vx, can be determined directly from the muzzle velocity:
Vx = V * cos(theta)
where V is the muzzle velocity and theta is the angle with respect to the horizontal.
2. Vertical motion:
The vertical component of velocity changes with time due to the acceleration due to gravity. We can determine the initial vertical velocity, Vy, and the time it takes for the cannonball to hit the ground.
The initial vertical velocity can be determined as:
Vy = V * sin(theta)
The time it takes for the cannonball to reach the ground can be found using the equation for vertical displacement:
y = ut + (1/2) * a * t^2
where y is the vertical displacement (in this case, -190 m, since the cannonball is fired downward), u is the initial vertical velocity (Vy), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Finally, the speed of the cannonball when it hits the ground can be determined using the horizontal and vertical components of velocity:
V = sqrt(Vx^2 + Vy^2)
Now you can use these equations to find the speed of the cannonball when it hits the ground.