A projectile is shot from the ground at a velocity of 4 m/s and at an angle of (5pi)/6. How long will it take for the projectile to land?

1 Answer
May 31, 2017

0.4 "s"

Explanation:

We're asked to find the time t when the projectile lands with a known initial velocity and direction. In terms of physics, we need to find the time when the change in y (the height) is equal to 0.

We can use the equation

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

to find the time t when Deltay is zero,

where

  • Deltay is the change in y-position (height) of the object, in "m"

  • v_(0y) is the initial y-velocity of the object, in "m"/"s"

  • t is the time, in "s", and

  • g is the acceleration due to gravity near Earth's surface, 9.80"m"/("s"^2).

To find the initial y-velocity, we can use the equation

v_(0y) = v_0sinalpha

where

  • v_0 is the magnitude of the initial velocity, in "m"/"s", and

  • alpha is the initial launch angle.

Plugging in our known variables, we have

v_(0y) = v_0sinalpha = (4"m"/"s")sin((5pi)/6) = color(red)(2"m"/"s"

Now that we have our known variables necessary, let's plug them back into the original equation to find the times when the height is 0 (there are two times because the height is zero when it is first launched as well):

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

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

1/2(9.80"m"/("s"^2))t^2 = (2"m"/"s")t

We'll eliminate the t = 0 option by dividing both sides by t:

(4.90"m"/("s"^2))t = (2"m"/"s")

t = (2cancel("m")/cancel("s"))/(4.90cancel("m")/("s"^cancel(2))) = color(blue)(0.4"s"

rounded to one significant figure, which is the amount given in the problem.

Thus, the object will land again ad t = 0.4"s"