# If a projectile is shot at an angle of #(5pi)/12# and at a velocity of #16 m/s#, when will it reach its maximum height??

1.545 seconds

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The projectile will reach its maximum height at the time when its vertical velocity component becomes zero. The time taken to reach maximum height can be calculated using the formula: ( t = \frac{v \cdot \sin(\theta)}{g} ), where ( v ) is the initial velocity (16 m/s), ( \theta ) is the launch angle ((5\pi/12)), and ( g ) is the acceleration due to gravity (9.8 m/s²).

Plugging in the values: ( t = \frac{16 \cdot \sin(5\pi/12)}{9.8} )

( t ≈ 1.03 ) seconds.

So, the projectile will reach its maximum height approximately 1.03 seconds after being launched.

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To find the time when the projectile reaches its maximum height, you can use the formula for the time taken to reach maximum height for a projectile launched at an angle:

[ t_{\text{max}} = \frac{V_0 \sin(\theta)}{g} ]

Where:

- ( V_0 ) is the initial velocity (16 m/s in this case)
- ( \theta ) is the launch angle (( \frac{5\pi}{12} ) in this case)
- ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 ) on the surface of the Earth)

Plug in the given values and calculate to find ( t_{\text{max}} ).

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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.

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