A projectile of mass 0.528 kg is shot from
a cannon, at height 6.7 m, as shown in the
figure, with an initial velocity vi having a
horizontal component of 7 m/s.
The projectile rises to a maximum height
of 3.8 m above the end of the cannon’s barrel
and strikes the ground a horizontal distance
∆x past the end of the cannon’s barrel.
∆x
vi
51
◦
3.8 m
6.7 m
Find the magnitude of the initial vertical
velocity vector of the cannon-ball at the end
of the cannon’s barrel. The acceleration of
gravity is 9.8 m/s
2
Answer in units of m/s.
Vo[51o]
Xo = 7 m/s
Vo = ?
Xo = Vo*Cos51 = 7 m/s
Vo = 7/Cos51 = 11.12 m/s = Initial velocity.
Yo = Vo*sin51 = 11.12*sin51 = 8.64 m/s.
To find the magnitude of the initial vertical velocity vector of the cannonball at the end of the cannon's barrel, we can use the equations of motion for projectile motion.
First, let's find the time it takes for the cannonball to reach its maximum height. We can use the equation:
H_max = (vi^2*sin^2θ) / (2*g)
where H_max is the maximum height, vi is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.
Using the given values, we have:
H_max = 3.8 m
vi = 7 m/s
θ = 51°
g = 9.8 m/s^2
Plugging in the values, we get:
3.8 = (7^2 * sin^2(51°)) / (2 * 9.8)
Now, we can rearrange the equation to solve for sin^2(51°):
sin^2(51°) = (3.8 * 2 * 9.8) / 7^2 = 0.244
Taking the square root, we get:
sin(51°) = √0.244 ≈ 0.494
The vertical component of velocity at the end of the cannon's barrel can be calculated using the equation:
v_y = vi * sin(θ)
Plugging in the values, we have:
v_y = 7 * 0.494 ≈ 3.459 m/s
Therefore, the magnitude of the initial vertical velocity vector of the cannonball at the end of the cannon's barrel is approximately 3.459 m/s.