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?
How long does it take to fall 40 m?
h=1/2 g t^2
solve for t.
then, velocity= 175/time
is g 9.8 or -9.8?
-9.8
Actually g is 9.8 if it were to be acceleration then it would be -9.8
9.80
To find the velocity with which the ball left the muzzle of the cannon, we can use the fact that the horizontal component of the velocity remains constant throughout the motion.
Given:
- Height of the tower (h) = 40.0 m
- Horizontal distance traveled by the cannonball (d) = 175 m
To solve this problem, we need to calculate the time it takes for the ball to reach the ground.
Step 1: Finding the time of flight (t)
We can use the equation of motion in the vertical direction to find the time of flight.
Using the equation: h = (1/2) * g * t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Plugging in the values, we get:
40 = (1/2) * 9.8 * t^2
Simplifying the equation gives us:
t^2 = (40 * 2) / 9.8
t^2 = 8.16
Taking the square root on both sides:
t = √8.16
t ≈ 2.86 seconds
Step 2: Finding the horizontal velocity (v_x)
Since the horizontal component of the velocity remains constant, we can directly calculate it using the formula:
v_x = d / t
v_x = 175 / 2.86
v_x ≈ 61.19 m/s
Therefore, the ball left the muzzle of the cannon with a velocity of approximately 61.19 m/s.