Assume you have a canon aimed straight up. The cannonball is shot at a velocity
of v_0=69 m/s.
How high does the cannonball go?
How much time does it take the cannonball to reach that height?
V^2 = Vo^2 + 2g*h.
0 = 69^2 - 19.6h, h = ?.
V = Vo + g*t.
0 = 69 - 9.8t, t = ?.
To find out how high the cannonball goes and how much time it takes to reach that height, we can use the basic equations of projectile motion. In this case, we assume there is no air resistance.
First, let's consider the vertical motion of the cannonball when shot straight up. The initial velocity (v₀) is 69 m/s, and we assume the acceleration due to gravity (g) is approximately 9.8 m/s² in the downward direction.
1. Time to reach maximum height:
When the cannonball reaches its maximum height, its vertical velocity (v) becomes zero. We can use the equation:
v = v₀ - gt
Rearranging the equation and solving for t, we get:
t = v₀ / g
Substituting the given values, we have:
t = 69 m/s / 9.8 m/s² ≈ 7.04 s
Therefore, it takes approximately 7.04 seconds for the cannonball to reach its maximum height.
2. Maximum height (h):
To find the maximum height, we can use the equation for displacement:
h = v₀t - (1/2)gt²
Since the ball is shot straight up, the displacement at maximum height is equal to zero. Thus, we can set h = 0 and solve for t:
0 = v₀t - (1/2)gt²
Rearranging the equation and solving for t, we get:
t² = 2v₀ / g
Substituting the given values, we have:
t² = 2 * 69 m/s / 9.8 m/s² ≈ 9.2 s²
Now we can substitute the value of t back into the equation for displacement to find h:
h = v₀t - (1/2)gt²
h ≈ 69 m/s * 7.04 s - 0.5 * 9.8 m/s² * 7.04 s²
Evaluating this equation, we find:
h ≈ 241.56 m
Therefore, the cannonball reaches a maximum height of approximately 241.56 meters.
To summarize:
- The cannonball takes approximately 7.04 seconds to reach its maximum height.
- The maximum height it reaches is approximately 241.56 meters.