A projectile of mass 0.413 kg is shot from a cannon, at height 6.8 m, with an initial velocity vi having a horizontal component of 7.5m/s.
The projectile rises to a maximum height of ∆y above the end of the cannon’s barrel and strikes the ground a horizontal distance ∆x past the end of the cannon’s barrel.
Determine the maximum height the projectile achieves after leaving the end of the cannon’s barrel.
Answer in units of m.
angle:
cosTheta=7.5/vi
time to fall 6.8m:
timein air=sqrt(6.5/g)
horizonal distance=7.5*timeinair
height:
mg*maxheight=1/2 m (vi*sinTheta)^2
solve for max height
To determine the maximum height the projectile achieves after leaving the end of the cannon's barrel, we can use the principles of projectile motion.
First, let's consider the vertical motion of the projectile. The initial vertical velocity (vi) is 0 since the projectile is shot horizontally. The only force acting on the projectile in the vertical direction is gravity, which causes it to accelerate downward with a magnitude of 9.8 m/s^2.
Using the equation for vertical displacement during constant acceleration:
∆y = vi * t + (1/2) * a * t^2
Since the initial vertical velocity is 0, the equation simplifies to:
∆y = (1/2) * a * t^2
Now, we need to find the time it takes for the projectile to reach its maximum height. At the highest point of its trajectory, the vertical velocity will be 0, so we can solve for t by setting the vertical velocity equation equal to 0:
vi + a * t = 0
Substituting the known values, we have:
0 + (-9.8) * t = 0
Solving for t:
t = 0
This means that the projectile reaches its highest point instantaneously, at the same time it leaves the end of the cannon's barrel.
Now, let's substitute this value of t in the equation for vertical displacement:
∆y = (1/2) * (-9.8) * (0^2)
∆y = 0
This tells us that the maximum height the projectile achieves after leaving the end of the cannon's barrel is 0 meters.
Therefore, the answer is 0 m.