Question

A cannonball was fired at a speed of 20 m/s at an angle of 35° from the ground with a range of 20 m. Find:

A) The maximum height reached by the cannonball.
B) The time of flight of the cannonball.

Answer 1

The maximum height reached by the cannonball is approximately 6.25 m, and the time of flight is approximately 4.10 s.

Step-by-step explanation:

To find the maximum height reached by the cannonball, we can use the equation for vertical motion. The formula for the maximum height is given by:

h = (v^2 * sin^2(theta))/(2g)

Where:

• h is the maximum height
• v is the initial velocity (20 m/s)
• theta is the launch angle (35°)
• g is the acceleration due to gravity (9.8 m/s^2)

Plugging in the values, we get:

h = (20^2 * sin^2(35°))/(2 * 9.8)

h ≈ 6.25 m

To find the time of flight of the cannonball, we can use the equation for the horizontal motion. The formula for the time of flight is given by:

t = 2 * (v * sin(theta))/g

Where:

• t is the time of flight
• v is the initial velocity (20 m/s)
• theta is the launch angle (35°)
• g is the acceleration due to gravity (9.8 m/s^2)

Plugging in the values, we get:

t = 2 * (20 * sin(35°))/9.8

t ≈ 4.10 s