A cannon ball is fired from ground level with a speed of 86.0 m/s at an angle of 32 deg. above the horizontal. A 8.5 m tall wall is located 265 m away from the cannon. By how much does the cannon ball clear the wall?

In the previous problem, what is the speed of the cannon ball when it is directly above the wall?

u = 86 cos 32 = 72.9 m/s forever

Vi = 86 sin 32 = 45.6 up at start

x distance = u t
265 = (86 cos 32) t
t = 3.63 seconds in the air
h = Hi + Vi t - 4.9 t^2
h = 0 + 45.6(3.63) -4.9(3.63)^2
h = 101 meters high

v = Vi - 9.81 t
at t = 3.63
v = 10 m/s up
u = 72.9 still
speed = sqrt (100 + 72.9^2)
about 73.6 m/s

To solve this problem, we can break it down into two parts: finding the maximum height reached by the cannonball and then finding the speed of the cannonball when it is directly above the wall.

First, let's find the maximum height reached by the cannonball. We can use the kinematic equation for vertical motion:

y = y0 + v0y * t - (1/2) * g * t^2

Where:
- y is the vertical displacement
- y0 is the initial vertical position (in this case, zero as the cannonball is fired from ground level)
- v0y is the vertical component of the initial velocity (v0 * sin(theta))
- t is the time taken to reach the maximum height
- g is the acceleration due to gravity (approximately 9.8 m/s^2)

Since the cannonball reaches its maximum height when its vertical velocity becomes zero, we can set v0y - g * t = 0. Solving for t, we find:

t = v0y / g

Now, let's find the vertical component of the initial velocity, v0y:
v0y = v0 * sin(theta)

Given:
- v0 = 86.0 m/s (speed of the cannonball)
- theta = 32 degrees (angle above the horizontal)

We can calculate v0y as follows:

v0y = 86.0 m/s * sin(32 deg)

Next, substitute v0y into the formula for time:

t = (86.0 m/s * sin(32 deg)) / 9.8 m/s^2

Now that we have the time taken to reach the maximum height, we can calculate the maximum height using the formula:

y = y0 + v0y * t - (1/2) * g * t^2

Since y0 = 0 (starting from ground level) and y is the maximum height, we have:

y = v0y * t - (1/2) * g * t^2

Substitute the values we have calculated:

y = (86.0 m/s * sin(32 deg)) * [(86.0 m/s * sin(32 deg)) / 9.8 m/s^2] - (1/2) * 9.8 m/s^2 * [(86.0 m/s * sin(32 deg)) / 9.8 m/s^2]^2

Now, let's move on to finding the speed of the cannonball when it is directly above the wall.

The horizontal component of the initial velocity, v0x, remains constant throughout the motion. We can find v0x using the formula:

v0x = v0 * cos(theta)

Given:
- v0 = 86.0 m/s (speed of the cannonball)
- theta = 32 degrees (angle above the horizontal)

We can calculate v0x as follows:

v0x = 86.0 m/s * cos(32 deg)

Now, let's find the time it takes for the cannonball to reach directly above the wall. The horizontal displacement, x, can be found using the formula:

x = x0 + v0x * t

Where:
- x0 is the initial horizontal position (in this case, 0 as the cannonball is fired from ground level)
- v0x is the horizontal component of the initial velocity
- t is the time taken to reach the wall

Given:
- x = 265 m (horizontal distance to the wall)
- x0 = 0 (starting from ground level)
- v0x = 86.0 m/s * cos(32 deg)

We can calculate t as follows:

265 m = 0 + (86.0 m/s * cos(32 deg)) * t

t = 265 m / (86.0 m/s * cos(32 deg))

Now that we have the time taken to reach directly above the wall, let's find the speed of the cannonball using the formula:

v = sqrt(vx^2 + vy^2)

Where:
- vx is the horizontal component of the velocity (v0x)
- vy is the vertical component of the velocity (v0y - g * t)

Given the values we have calculated, substitute them into the formula:

v = sqrt((86.0 m/s * cos(32 deg))^2 + ((86.0 m/s * sin(32 deg)) - 9.8 m/s^2 * t)^2)

Simplify the equation and calculate the value of v using the given values. This will give us the speed of the cannonball when it is directly above the wall.