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?

1 Answer
Oct 14, 2016

Answer:

#t ~~ 10.6 s#

Explanation:

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#