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

Answer 1

Layla Ahmed

#0.7"s"#

Apply the motion equation:

#v=u+at#

The vertical component of the initial velocity is #vsintheta#.

We can calculate the time required to reach the maximum height because the final velocity is zero at this point.

This is where the equation changes to:

#0=vsintheta-"g"t#

#:.t=(vsintheta)/g#

Radians to degrees conversion:

#pi=180^@#

#:.pi/3=180^@/3=60^@#

#:.t=4sin60/9.8=0.35"s"#

The second leg of the flight is when the projectile falls to earth. Because the path is symmetrical this will also be equal to #0.35"s"#.

So total time of flight = #0.35xx2=0.7"s"#

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Answer 2

Olivia Askew

To find the time it takes for the projectile to land, you can use the formula ( t = \frac{2v_0 \sin(\theta)}{g} ), where ( v_0 ) is the initial velocity, ( \theta ) is the launch angle, and ( g ) is the acceleration due to gravity. Substituting the given values, ( v_0 = 4 , \text{m/s} ), ( \theta = \frac{\pi}{3} ), and ( g = 9.8 , \text{m/s}^2 ), you get ( t = \frac{2 \times 4 \times \sin(\frac{\pi}{3})}{9.8} ). Solve for ( t ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.