Question

A cannon is mounted on a tower above a wide, level field. The barrel of the cannon is 20 m above the ground below. A cannonball is fired horizontally with an initial speed of 700 m/s. Assuming air resistance can be neglected, approximately how long will the cannonball be in flight before it hits the ground?

Answer 1

The question discusses Projectile Motion in Physics, asking how long a horizontally fired cannonball at 20 m height will be in the air before hitting the ground. The time is found by solving a kinematics equation for vertical motion, resulting in approximately 2.02 seconds.

Step-by-step explanation:

The subject of the student's question is Projectile Motion, a topic within Physics. Since the cannonball is fired horizontally from a height of 20 meters, we can neglect the horizontal motion when determining the time it takes to hit the ground. This is a straightforward kinematics problem where we only need to consider the vertical motion due to gravity.

To find the time of flight, we use the equation for the motion under constant acceleration (gravity), which is:

Distance (d) = Initial velocity (vi) × Time (t) + ½ × Acceleration (a) × Time2

Since the cannonball is fired horizontally, the initial vertical velocity is 0 m/s (vi = 0). The acceleration is due to gravity, which is approximately 9.81 m/s2 (a = 9.81 m/s2). Plugging in the values:

20 = 0 × t + ½ × 9.81 m/s2 × t2

Solving for time (t), we will get approximately 2.02 seconds.

Answer 2

Cannonball will be in flight before it hits the ground for 2.02 seconds

Step-by-step explanation:

Initial height from ground = 20 meter.

We have equation of motion ,
`s= ut+(1)/(2) at^2`, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8
`m/s^2`, we need to calculate time when s = 20 meter.

Substituting

`20=0*t+(1)/(2) *9.8*t^2\\ \\ t = 2.02 seconds`

So it will take 2.02 seconds to reach ground.