A projectile is shot from the ground at a velocity of #32 m/s# and at an angle of #(3pi)/4#. How long will it take for the projectile to land?

To find the time it takes for the projectile to land, you can use the following equation: time = (2 * initial velocity * sine of the launch angle) / acceleration due to gravity. Plugging in the given values: initial velocity = 32 m/s, launch angle = (3π)/4, and acceleration due to gravity = 9.8 m/s², the calculation yields a time of approximately 4.10 seconds.

To determine how long it will take for the projectile to land, we need to use the kinematic equations of motion. The horizontal and vertical motion can be treated separately.

First, let's break down the initial velocity into its horizontal and vertical components. The given velocity of 32 m/s at an angle of ( \frac{3\pi}{4} ) can be expressed as:

Horizontal component: ( v_{0x} = v_0 \cdot \cos(\theta) = 32 \cdot \cos\left(\frac{3\pi}{4}\right) = -\frac{32}{\sqrt{2}} = -16\sqrt{2} ) m/s Vertical component: ( v_{0y} = v_0 \cdot \sin(\theta) = 32 \cdot \sin\left(\frac{3\pi}{4}\right) = \frac{32}{\sqrt{2}} = 16\sqrt{2} ) m/s

The negative sign in the horizontal component indicates motion in the negative x-direction.

Next, we can use the vertical motion equation ( y = y_0 + v_{0y}t + \frac{1}{2}at^2 ), where ( y_0 = 0 ) m (starting from the ground), ( a = -9.8 ) m/s² (acceleration due to gravity acting downward), and ( y ) is the vertical distance traveled.

Let's solve for time (( t )) when the projectile lands, which means ( y = 0 ):

[ 0 = 0 + (16\sqrt{2})t + \frac{1}{2}(-9.8)t^2 ]

Now, solve this quadratic equation for ( t ):

[ -4.9t^2 + 16\sqrt{2}t = 0 ]

Factor out ( t ):

[ t(-4.9t + 16\sqrt{2}) = 0 ]

This equation gives two possible solutions: ( t = 0 ) (the initial time, which we discard) and ( t = \frac{16\sqrt{2}}{4.9} ).

Calculate the time:

[ t = \frac{16\sqrt{2}}{4.9} \approx 2.06 ] seconds

So, it will take approximately 2.06 seconds for the projectile to land.