Question

A melon is dropped from rest from the top of a 30 meter cliff. At the same time, a person at the bottom of the cliff shoots a paintball up towards the melon. The paintball strikes the melon after 0.2 seconds. (a) How far from the base of the cliff does the paintball strike the melon? (b) What was the initial velocity of the paintball?

Answer 1

Let's solve the problem step by step:

Given information:

• Initial height (h) of the melon = 30 meters
• Time taken for the paintball to strike the melon (t) = 0.2 seconds

(a) To find the horizontal distance from the base of the cliff where the paintball strikes the melon, we can use the formula for horizontal distance traveled by an object in free fall:

• 
  `Horizontal`
  `Distance=`
  `Horizontal`
  `Velocity`
  `*`
  `time`

Since the paintball and the melon are both falling under gravity, their horizontal velocities are the same.

• Horizontal velocity(
  `v_(x)`)= ? (This is what we need to find)
• Given that the time(
  `t\\`) is 0.2 seconds.

We know that the horizontal distance covered by the object is equal to the horizontal velocity times the time of flight:

• 
  `Horizontal`
  `Distance=`
  `v_(x) *t`

Since the paintball hits the melon after 0.2 seconds, we can use this time to calculate the horizontal distance.

(b) The initial vertical velocity (
`v_(y0)`) of the paintball can be determined using the equation for free fall motion:

• 
  `v_(y) =v_(y0)-g . t`

Where:

• 
  `v_(y)` is the final vertical velocity (which is 0 because the paintball reaches the melon)
• 
  `g` is the acceleration due to gravity (approximately 9.81 m/s²)
• 
  `t` is the time of flight, which is 0.2 seconds

Solving for
`v_(y0)`:

• 
  `v_(y0)=v_(y) -g . t`

Now that we have
`v_(y0)`, we can use it to find the initial velocity (
`v_(0)`) of the paintball. The initial velocity is the vector sum of the horizontal and vertical components:

• 
  `v_(0) = \sqrt{v2_(x0) + {v2_(y0)}`

Let's calculate:

(a) Horizontal distance:

• 
  `Horizontal`
  `Distance = v_(x)` *
  `t`

(b) Initial vertical velocity:

• 
  `v_(y0)=v_(y) +g . t`

(c) Initial velocity:

• 
  `v_(0) = \sqrt{v2_(x) + {v2_(y0)}`

Remember that the horizontal velocity (
`v_(x)`) is the same for both the paintball and the melon.