Question

A dart is thrown at a dartboard 3.66 m away. When the dart is released at the same height as the center of the dartboard, it hits the center in 0.455 s. (Neglect any effects due to air resistance.)At what angle relative to the floor was the dart thrown?

Answer 1

The angle is
`\theta = 15.48^o`

Step-by-step explanation:

From the question we are told that

The distance of the dartboard from the dart is
`d = 3.66 \ m`

The time taken is
`t = 0.455 \ s`

The horizontal component of the speed of the dart is mathematically represented as

`u_x = ucos \theta`

where u is the the velocity at dart is lunched

so

`distance = velocity \ in \ the\ x-direction * time`

substituting values

`3.66 = ucos \theta * (0.455)`

=>
`ucos \theta = 8.04 \ m/s`

From projectile kinematics the time taken by the dart can be mathematically represented as

`t = (2usin \theta )/(g)`

=>
`usin \theta = (g * t)/(2 )`

`usin \theta = (9.8 * 0.455)/(2 )`

`usin \theta = 2.23`

=>
`tan \theta = (usin\theta )/(ucos \theta ) = (2.23)/(8.04)`

`\theta = tan^(-1) [0.277]`

`\theta = 15.48^o`