A cannonball is shot (from ground level) with an initial horizontal velocity of 36.0 m/s and an initial vertical velocity of 21.0 m/s.

1) What is the initial speed of the cannonball?
2) What is the initial angle θ of the cannonball with respect to the ground?
3) What is the maximum height the cannonball goes above the ground?
4) How far from where it was shot will the cannonball land?
5) What is the speed of the cannonball 2.7 seconds after it was shot?
6) How high above the ground is the cannonball 2.7 seconds after it is shot?

1) Use the Pythagorean theorem.

Initial Speed = Vo = sqrt [Vx^2 + Vy^2]
2) Launch Angle = arctan(Vyo/Vx)
3) g*Hmax = Vyo^2/2
where Vyo is the initial vertical velocity component.
4) Range = 2*(Vo^2/g)*sinAcosA
5) Use the Pythagorean theorem again, but with the changed value of Vy.
Vy = Vyo - g*t
6) Height = Vyo*t - (g/2)*t^2

To solve these questions, we can use the principles of projectile motion and break down the motion into horizontal and vertical components. Let's go step by step:

1) The initial speed (also known as the magnitude of the initial velocity) of the cannonball can be found using the Pythagorean theorem:

Initial speed = √(initial horizontal velocity^2 + initial vertical velocity^2)
Initial speed = √(36.0 m/s)^2 + (21.0 m/s)^2)
Initial speed = √(1296 + 441)
Initial speed = √1737
Initial speed ≈ 41.67 m/s

2) The initial angle θ of the cannonball with respect to the ground can be found using trigonometry:

tan(θ) = (initial vertical velocity) / (initial horizontal velocity)
θ = arctan((21.0 m/s) / (36.0 m/s))
θ ≈ 30.96°

3) The maximum height the cannonball goes above the ground can be found using the formula for vertical displacement:

Maximum height = (initial vertical velocity)^2 / (2 * acceleration due to gravity)
Maximum height = (21.0 m/s)^2 / (2 * 9.8 m/s^2)
Maximum height ≈ 22.24 m

4) The horizontal distance the cannonball travels can be found using the formula for horizontal displacement and time of flight:

Horizontal distance = (initial horizontal velocity) * (time of flight)
Horizontal distance = (36.0 m/s) * (2 * (initial vertical velocity) / (acceleration due to gravity))
Horizontal distance = (36.0 m/s) * (2 * 21.0 m/s) / 9.8 m/s^2)
Horizontal distance ≈ 149.94 m

5) To find the speed of the cannonball 2.7 seconds after it was shot, we can use the horizontal and vertical components of velocity:

Speed after 2.7 seconds = √((horizontal velocity)^2 + (vertical velocity 0 - (acceleration due to gravity) * time)^2)
Speed after 2.7 seconds = √((36.0 m/s)^2 + (21.0 m/s - (9.8 m/s^2) * 2.7 s)^2)

6) To find the height above the ground after 2.7 seconds, we can use the formula for vertical displacement:

Height after 2.7 seconds = (initial vertical velocity) * (time) - (0.5 * acceleration due to gravity * time^2)
Height after 2.7 seconds = (21.0 m/s) * (2.7 s) - (0.5 * 9.8 m/s^2 * (2.7 s)^2)

To solve these questions, we can use the principles of projectile motion. Let's break down each question step by step:

1) The initial speed of the cannonball can be found using the Pythagorean theorem. The horizontal and vertical components of the initial velocity (Vx and Vy) are given as 36.0 m/s and 21.0 m/s, respectively. Therefore, the initial speed can be calculated as follows:

Initial speed = √(Vx^2 + Vy^2)
Initial speed = √(36.0^2 + 21.0^2) m/s

2) The initial angle (θ) of the cannonball with respect to the ground can be determined using trigonometry. We can use the inverse tangent function (arctan) to find the angle. The angle θ can be calculated as follows:

θ = arctan(Vy / Vx)
θ = arctan(21.0 / 36.0) degrees

3) The maximum height the cannonball reaches can be determined by calculating the vertical displacement. We can use the kinematic equation for vertical motion:

Vertical displacement = (Vy^2) / (2 * acceleration due to gravity)
Vertical displacement = (21.0^2) / (2 * 9.8) meters

4) The horizontal distance the cannonball travels can be calculated using the formula:

Horizontal distance = Vx * time of flight
Horizontal distance = 36.0 * time of flight meters

To find the time of flight, we can use the equation:

time of flight = (2 * Vy) / (acceleration due to gravity)
time of flight = (2 * 21.0) / 9.8 seconds

5) To find the speed of the cannonball 2.7 seconds after it was shot, we need to calculate its horizontal and vertical components at that time. The horizontal component remains constant, but the vertical component changes due to the effect of gravity. The speed can be calculated using the Pythagorean theorem:

Speed = √(Vx^2 + (Vy - gravity * time)^2)
Speed = √(36.0^2 + (21.0 - 9.8 * 2.7)^2) m/s

6) To determine the height of the cannonball above the ground after 2.7 seconds, we can use the equation for vertical displacement:

Vertical displacement = (Vy * time) + (0.5 * acceleration due to gravity * time^2)
Vertical displacement = (21.0 * 2.7) + (0.5 * 9.8 * (2.7)^2) meters

By plugging in the given values and performing the calculations, you can find the answers to each question.