A projectile is shot from the ground at a velocity of #17 m/s# and at an angle of #(7pi)/12#. How long will it take for the projectile to land?

1 Answer
May 31, 2017

Answer:

#3.3# #"s"#

Explanation:

We're asked to find the time when the projectile lands after being launched with a known initial launch velocity (magnitude and direction. In terms of physics, we need to find the time #t# when the change in #y#-position #Deltay# (the height) is equal to zero.

To solve this, we can use the equation

#Deltay = v_(0y)t - 1/2g t^2#

Knowing that #Deltay# is #0#, and #g# is #9.80"m"/("s"^2)#, we still need to know the initial #y#-velocity #v_(0y)#, which we can find by

#v_(0y) = v_0sinalpha = (17"m"/"s")sin((7pi)/12) = color(red)(16.4 "m"/"s"#

Therefore, plugging in known variables, we have

#0 = (color(red)(16.4"m"/"s"))t - 1/2(9.80"m"/("s"^2))t^2#

#(4.90"m"/("s"^2))t = color(red)(16.4"m"/"s"#

#t = color(blue)(3.3"s"#

rounded to #2# significant figures, the amount given to us in the problem.

Therefore, the object will land after #3.3# #"s"#.

Something we could do (for future reference and for fun) is to find
a general expression for the time #t# for any launched projectile when its height is the same as its launch height (zero, in this case).

Let's write our original equation but leave all the variables un-substituted, except for #Deltay#, which is #0#:

#0= v_(0y)t - 1/2g t^2#

What we can first do is divide both sides by #t# to leave just one remaining (by doing this, we're nullifying the #t = 0# solution, which is merely pointless for most calculations:

#0= v_(0y) - 1/2g t#

Now, let's move the acceleration group to the left side:

#1/2g t = v_(0y)#

Multiplying both sides by #2# (dividing by #1/2#):

#g t = 2v_(0y)#

And isolating #t#, we have

#t = (2v_(0y))/g#

Thus, whenever a projectile is launched from any height, the time #t# when it reaches that height again is given by the equation #t = (2v_(0y))/g#

Intuitively, the projectile will only reach the same height again if it is launched (at least somewhat) upward; that is, it's initial #y#-velocity is positive.

We can check this from the equation: if #v_(0y)# is #0# (it's launched horizontally), then the time is also #0#; it is only at that height once in its motion, even before it was launched (the motion equation, if graphed, will show that this height is its maximum height; that is, if the projectile was launched at a time before it was in your hand, it would reach your hand at this maximum height).

If #v_(0y)# is negative, it was launched (at least somewhat) downward, and the time given will be negative, which makes no sense realistically, but theoretically, if it continued the motion by which it was projected before it was launched, it would be at that height at that calculated time, before it was even launched.

Lastly, let's check to see if this general equation works (if it gives us the same time):

#t = (2(16.4cancel("m")/"s"))/(9.80cancel("m")/(cancel("s"^2))) = color(blue)(3.3"s"#

Indeed it does:)