A projectile is shot from the ground at a velocity of #5 m/s# and at an angle of #(7pi)/12#. How long will it take for the projectile to land?

Answer 1

#0,9856s#

The only force acting on the projectile after being fired is the force of gravity.

We may resolve the motion into 2 components :

In the horizontal (X) direction, there is constant velocity of #5cos(7pi)/12rad=5cos105^@=5cos75^@1,2941m//2#
Note that we take the angle with the horizontal axis as the reference angle and so the projectile is actually being fired backwards with a projection angle of #75^@#.

In the vertical (Y) direction, there is constant acceleration due to gravity and hence the equations of motion for uniform acceleration in 1 direction are valid.

The initial velocity in the y-direction is hence #5sin75^@=4,8296m//s#.
Therefore, time taken to reach maximum height is given by : #v=u+at => t=(v-u)/a# #therefore t=(0-4,8296)/(-9,8)=0,4928s#.
Therefore the total time taken to reach ground again, since the path is symmetric, is #0,4928xx2=0,9856s#.
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Answer 2

To find the time it takes for the projectile to land, we can use the formula for the time of flight of a projectile:

[ t = \frac{2v_0 \sin(\theta)}{g} ]

where:

  • ( v_0 ) is the initial velocity (5 m/s in this case),
  • ( \theta ) is the launch angle (( \frac{7\pi}{12} ) radians in this case),
  • ( g ) is the acceleration due to gravity (approximately 9.8 m/s²).

Plugging in the values:

[ t = \frac{2 \times 5 \times \sin\left(\frac{7\pi}{12}\right)}{9.8} ]

[ t \approx \frac{10 \times 0.9659}{9.8} ]

[ t \approx \frac{9.659}{9.8} ]

[ t \approx 0.9855 \text{ seconds} ]

So, it will take approximately 0.9855 seconds for the projectile to land.

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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.

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.

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.

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.

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