# A projectile is shot from the ground at an angle of #( pi)/3 # and a speed of #9 /4 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

Applying the equation of motion

Therefore,

Applying the equation of motion

The distance travelled in the horizontal direction is

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 can use the equation for the horizontal distance traveled:

[ \text{Horizontal distance} = \text{initial horizontal velocity} \times \text{time} ]

Since the projectile reaches its maximum height halfway through its flight, the time it takes to reach maximum height can be found using the equation for vertical motion:

[ \text{Time to maximum height} = \frac{\text{initial vertical velocity}}{\text{gravitational acceleration}} ]

Given that the initial vertical velocity is ( V_{i_y} = V \sin(\theta) ), where ( V ) is the initial velocity and ( \theta ) is the launch angle, and gravitational acceleration is ( g = 9.8 , \text{m/s}^2 ), we can plug in the values:

[ V_{i_y} = \frac{9}{4} \times \sin\left(\frac{\pi}{3}\right) = \frac{9}{4} \times \frac{\sqrt{3}}{2} = \frac{9\sqrt{3}}{8} ]

[ \text{Time to maximum height} = \frac{\frac{9\sqrt{3}}{8}}{9.8} = \frac{9\sqrt{3}}{78} ]

Since the projectile reaches maximum height halfway through its flight, the total time of flight is twice the time to maximum height:

[ \text{Total time of flight} = 2 \times \frac{9\sqrt{3}}{78} = \frac{9\sqrt{3}}{39} ]

Now, we can use the total time of flight to find the horizontal distance traveled:

[ \text{Horizontal distance} = \text{initial horizontal velocity} \times \text{total time of flight} ]

[ \text{Horizontal distance} = \frac{9}{4} \times \cos\left(\frac{\pi}{3}\right) \times \frac{9\sqrt{3}}{39} ]

[ \text{Horizontal distance} = \frac{9}{4} \times \frac{1}{2} \times \frac{9\sqrt{3}}{39} = \frac{81\sqrt{3}}{156} ]

Therefore, the horizontal distance from the starting point when the projectile reaches its maximum height is ( \frac{81\sqrt{3}}{156} ) meters.

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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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