Blog: A dart is projected upwards at an angel of 20° with the horizontal. It has an initial velocity of 20 m/s. In how many seconds will it reach the...

Sunday, February 24, 2013

I am assuming (in absence of any further information in the question) that the dart board is at the same vertical level as the point of projection.

Given data: initial velocity, u = 20 m/s

angle of projection, $\displaystyle\theta$ = 20

Using the equation of projectile motion:

$\displaystyle{y}={u}_{{y}}{t}-\frac{{1}}{{2}}{{g{{t}}}}^{{{2}}}$

where, $\displaystyle{u}_{{y}}$ is vertical component of velocity and is given as u sin( $\displaystyle\theta$ ).

Substituting y = 0, we get,

t = 0 sec (initial instant) and

t = $\displaystyle\frac{{{2}{u}{\sin{\theta}}}}{{{g}}}={1.395}{\sec{}}$

Thus, it would take about 1.4 seconds for the dart to hit the board.

The point of projection is at a distance equal to range of the projectile motion. Using the equation for range,

$\displaystyle{R}=\frac{{{{u}}^{{2}}{\sin{{2}}}\theta}}{{{g}}}=\frac{{{{20}}^{{2}}{\sin{{\left({2}\times{20}\right)}}}}}{{{9.81}}}={26.21}{m}$

Thus, the dart will strike its target 26.21 m from the point of projection.

Hope this helps.

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