A projectile is shot from the ground at a velocity of $54 m/s$ and at an angle of $(7pi)/12$ . How long will it take for the projectile to land?

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Answer 1 · 2016-10-14T17:00:06

$t ~~ 10.6 s$

Write the equation for the y coordinate as a function of time:

$y(t) = (54 m/s)sin((7pi)/12)t - (1/2)(9.8 m/s^2)t^2$

To find the times when the projectile is on the ground we set $y(t) = 0$ :

$0 = (54 m/s)sin((7pi)/12)t - (1/2)(9.8 m/s^2)t^2$

$t((54 m/s)sin((7pi)/12) - (1/2)(9.8 m/s^2)t) = 0$

We know that the projectile is launched from the ground so $t = 0$ is an extraneous root and, therefore, we can divide the equation by t:

$(54 m/s)sin((7pi)/12) - (1/2)(9.8 m/s^2)t = 0$

Solve the remaining factor for t:

$t = 2(54 m/s)/(9.8 m/s^2)sin((7pi)/12)$

$t ~~ 10.6 s$

A projectile is shot from the ground at a velocity of 54 m/s54 m/s and at an angle of (7pi)/12(7pi)/12. How long will it take for the projectile to land?