# A projectile is shot from the ground at an angle of #pi/4 # and a speed of #3/2 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

The distance is

We apply the equation of motion

to calculate the time to reach the greatest height

The greatest height is

To find the horizontal distance, we apply the equation of motion

The distance from the starting point is

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To find the distance from the starting point when the projectile reaches its maximum height, we need to calculate the horizontal component of the projectile's motion. The horizontal component of velocity remains constant throughout the projectile's flight. We can use the formula for horizontal motion:

[ \text{Horizontal distance} = \text{Horizontal velocity} \times \text{Time} ]

Given that the angle of projection is ( \frac{\pi}{4} ) radians and the speed of the projectile is ( \frac{3}{2} ) m/s, the horizontal component of the velocity can be found using trigonometric functions:

[ \text{Horizontal velocity} = \text{Initial velocity} \times \cos(\text{angle of projection}) ]

[ \text{Horizontal velocity} = \frac{3}{2} \times \cos\left(\frac{\pi}{4}\right) = \frac{3}{2} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4} ]

Since the vertical component of velocity is zero at the maximum height, we can use the vertical motion equation to find the time it takes to reach maximum height:

[ \text{Vertical velocity} = \text{Initial velocity} \times \sin(\text{angle of projection}) ]

[ \text{Vertical velocity} = \frac{3}{2} \times \sin\left(\frac{\pi}{4}\right) = \frac{3}{2} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4} ]

Then, we use the formula for time of flight:

[ \text{Time to reach maximum height} = \frac{\text{Vertical velocity}}{\text{Gravity}} ]

[ \text{Time to reach maximum height} = \frac{\frac{3\sqrt{2}}{4}}{9.8} ]

[ \text{Time to reach maximum height} \approx \frac{3\sqrt{2}}{39.2} ]

Now, we can use this time to calculate the horizontal distance:

[ \text{Horizontal distance} = \frac{3\sqrt{2}}{4} \times \frac{3\sqrt{2}}{39.2} ]

[ \text{Horizontal distance} \approx \frac{18}{39.2} ]

[ \text{Horizontal distance} \approx 0.459 \text{ meters} ]

Therefore, when the projectile reaches its maximum height, it will be approximately 0.459 meters away from the starting point horizontally.

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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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