Answer:
a) The maximum height of the cannonball is 64.5 m.
b) The cannonball´s speed at maximum height is 50.8 m/s.
Explanation:
The position and velocity vectors of the cannonball can be calculated using the following equations:
r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)
v = (v0 · cos α, v0 · sin α + g · t)
Where:
r = position vector at time t.
x0 = initial horizontal position.
v0 = initial velocity.
t = time.
α = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive).
v = velocity vector at time t.
a) At the maximum height, the vertical component of the velocity vector is 0 (please, see the attached figure and notice that at the maximum height the velocity vector is horizontal).
Knowing this, we can calculate the time at which the cannonball is at its maximum height:
vy = v0 · sin α + g · t
0 m/s = 62.0 m/s · sin 35.0° - 9.81 m/s² · t
- 62.0 m/s · sin 35.0° / -9.81 m/s² = t
t = 3.63 s
Now, we can calculate the y-component of the vector r1 in the figure (r1y):
y = y0 + v0 · t · sin α + 1/2 · g · t²
The cannon is at the same level that the origin of the frame of reference (the ground) so that y0 = 0.
y = 0 m + 62.0 m/s · 3.63 s · sin 35.0° - 1/2 · 9.81 m/s² · (3.63 s)²
y = 64.5 m
The maximum height of the cannonball is 64.5 m
b) To calculate the speed at the maximum height, we can use the equation of the velocity vector:
v = (v0 · cos α, v0 · sin α + g · t)
We already know that the y-component is 0. Then, let´s calculate the x-component of the velocity:
vx = v0 · cos α
vx = 62.0 m/s · cos 35.0°
vx = 50.8 m/s
The vector velocity at maximum height will be:
v = (50.8 m/s, 0)
The speed is the magnitude of the velocity vector:
[tex]|v| = \sqrt{(50.8 m/s)^{2} + (0 m/s)^{2}} = 50.8 m/s[/tex]
The cannonball´s speed at maximum height is 50.8 m/s.
