A cannon is fired horizontally from atop a 40.0 m tower. The cannonball travels 175 m horizontally before it strikes the ground. With what velocity did the ball leave the muzzle?
Vx = (175 m)/(time to fall)
Time to fall = sqrt(2 H/g) = 2.86 s
Solve for Vx
how did u calculate "time to fall". and what is H and g in the equation:
sqrt(2 H/g) = 2.86 s?
To find the initial velocity of the ball, we can use the equations of motion. Since the cannonball was fired horizontally, we know that its vertical velocity is 0 m/s when it leaves the muzzle.
We can use the following equation to find the time it takes for the cannonball to reach the ground:
h = (1/2) * g * t^2
Where h is the height of the tower (40.0 m), g is the acceleration due to gravity (9.8 m/s^2), and t is the time taken.
Rearranging the equation, we have:
t = sqrt((2 * h) / g)
Plugging in the values, we get:
t = sqrt((2 * 40.0) / 9.8)
t ≈ 2.02 seconds
Now that we have the time, we can find the horizontal velocity using the equation:
v = d / t
Where v is the velocity, d is the horizontal distance traveled (175 m), and t is the time taken (2.02 s).
v = 175 / 2.02 ≈ 86.63 m/s
Therefore, the ball left the muzzle with a velocity of approximately 86.63 m/s.