Question

You are woken up in the morning on the farm by a rooster. The rooster is located 30 meters away on top of the 5 meter barn house. If you throw your alarm clock at an angle of 53 degrees at the rooster, so that it hits the rooster on the way up, with what velocity do you need to throw your alarm clock

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is
`v = 18.73 \ m/s`

Step-by-step explanation:

From the question we are told that

The distance of the rooster from farm house is
`d = 30 \ m`

The height of the rooster from the ground is
`h = 5 \ m`

The angle at which the alarm clock is thrown is
`\theta = 53 ^o`

Let the velocity at which the alarm clock is thrown be v

So the horizontal component of v is mathematically represented as

`v_h = v cos \theta`

The distance covered by the alarm clock toward the horizontal direction at velocity v is

`d = v_h`

=>
`d = v_h = vcos \theta * t`

=>
`vcos \theta * t = 30`

substituting for
`\theta`

`v (cos (53)) * t = 30`

=>
`v * t = 50`

Considering the motion of the alarm clock in the vertical direction

So the vertical component of v is mathematically represented as

`v_v = vsin \theta`

The distance covered in the vertical direction is mathematically evaluated as follows

From the equation of motion we have

`h = vsin\theta * t - (1)/(2) g t^2`

`v* t sin\theta - 4.9t^2 = 5`

Recall
`v * t = 50`

So

`50 \ sin \theta -4.9t^2 = 5`

substituting for
`\theta`

`50 \ sin (53) -4.9t^2 = 5`

=>
`t =√(7.1289)`

`t =2.673 \ s`

Now

`v * t = 50`

So

`v * 2.673 = 50`

=>
`v = 18.73 \ m/s`