Question

A 19.0-kg cannonball is fired from a cannon with muzzle speed of 800 m/s at an angle of 33.0° with the horizontal. A second ball is fired with the same initial speed at an angle of 90.0°. Let y = 0 at the cannon.Use the isolated system model to find the maximum height reached by each ball. hfirst ball = _________.

Answer 1

The maximum height reached by each cannonball is calculated using the initial vertical component of the velocity and conservation of energy, which transforms kinetic energy into gravitational potential energy at the peak of the trajectory. For the first ball fired at an angle, trigonometry is used to find the initial vertical velocity component. For the second ball fired straight up, its initial velocity is its muzzle speed.

Step-by-step explanation:

To find the maximum height reached by each cannonball, we will use the conservation of energy. For the first ball fired at a 33.0° angle, we only need to focus on the vertical component of the velocity, since only that influences the height it will reach. The initial kinetic energy in the vertical direction is transformed into gravitational potential energy at the maximum height.

The initial vertical velocity (Vy) is given by:

Vy = V × sin(θ)

Where V is the muzzle speed (800 m/s) and θ is the launch angle (33.0°).

For the first ball:

Vy = 800 m/s × sin(33.0°)

Using conservation of energy, where kinetic energy (KE) turns into potential energy (PE), we get:

PE at highest point = KE initial vertical

We will use the formula:

PE = m × g × h

For the second ball fired vertically, the calculation is simpler as its initial vertical speed is its muzzle speed.

By setting up the equations and solving for h (maximum height), we can find the answer for each ball.

Answer 2

The maximum height reached by both balls are 9685.93 meters and 32653.06 meters respectively.

Step-by-step explanation:

Given that,

Mass of the first cannonball,
`m_1=19\ kg`

Initial speed of the first cannonball,
`u_1=800\ m/s`

Angle of projection of the first cannonball,
`\theta_1=33^(\circ)`

Angle of projection of the second cannonball,
`\theta_2=90^(\circ)`

In this case, we need to find the maximum height reached by each ball. We know that the maximum height reached by a projectile is given by :

`h_1=((u\ \sin\theta_1)^2)/(2g)\\\\h_1=((800\ \sin(33))^2)/(2* 9.8)\\\\h_1=9685.93\ m`

The maximum height reached by the second ball is given by :

`h_2=((u\ \sin\theta_2)^2)/(2g)\\\\h_2=((800\ \sin(90))^2)/(2* 9.8)\\\\h_2=32653.06\ m`

So, the maximum height reached by both balls are 9685.93 meters and 32653.06 meters respectively. Hence, this is the required solution.