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