Question

A dart is thrown toward the target 3.2m away. The dart is thrown with a velocity of10.9m/s at an angle of 7.59 from the ground. Does the dart hit the target?

Answer 1

Given data:

* The initial velocity of the dart is 10.9 m/s.

* The angle of the initial velocity with the ground is 7.59 degree.

Solution:

The time period of the projectile motion is,

`t=(2u\sin(\theta))/(g)`

where u is the initial velocity, g is the acceleration due to gravity, and

`\theta\text{ is the angle with the ground,}`

Substituting the known values,

`\begin{gathered} t=(2*10.9*\sin(7.59))/(9.8) \\ t=0.29\text{ s} \end{gathered}`

By the kinematics equation, the horizontal range of the dart is,

`R=u_xt+(1)/(2)at^2`

where u_x is the initial horizontal velocity, a is the acceleration, and t is the time of flight,

The acceleration of the dart is zero along the horizontal direction,

`\begin{gathered} R=10.9\cos (7.59)*0.2938 \\ R=3.17 \\ R\approx3.2\text{ m} \end{gathered}`

Thus, the distance traveled by the dart in the horizontal direction is 3.2 meter.

Hence, the dart can hit the target.