# A projectile is shot from the ground at an angle of #pi/4 # and a speed of #2/5 ms^-1#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

For Physics or Mechanics you should learn the "suvat" equations for motion under constant acceleration:

Vertical Motion

Motion under constant acceleration due to gravity, applied vertically upwards

Let the total time that the projectile takes to its maximum point be

# { (s=,"not required",m),(u=,2/5 sin(pi/4)=1/5sqrt(2),ms^-1),(v=,0,ms^-1),(a=,-g,ms^-2),(t=,T,s) :} #

So we can calculate

# :. 0=1/5sqrt(2)-gT #

# :. gT=1/5sqrt(2) #

# :. T=sqrt(2)/(5g)#

Horizontal Motion

Under constant speed (NB we can still use "suvat" equation with a=0). Thje projectile will be in in the air for the same time,

So we can calculate

# s=2/5cos(pi/4)T #

# :. s=1/5sqrt(2)T #

# :. s=1/5sqrt(2)sqrt(2)/(5g) #

# :. s=2/(25g) #

So using

#s=2/245 = 0.008163...= 0.0081 m# , or#8.1mm#

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To find the distance from the starting point when the projectile reaches its maximum height, you can use the formula:

[ d = v_{0x} \cdot t ]

where ( v_{0x} ) is the initial velocity in the horizontal direction and ( t ) is the time it takes for the projectile to reach its maximum height. The horizontal component of the initial velocity can be calculated as ( v_{0x} = v_0 \cdot \cos(\theta) ), where ( v_0 ) is the initial speed and ( \theta ) is the angle of projection.

Given:

- Initial speed (( v_0 )) = ( \frac{2}{5} ) m/s
- Angle of projection (( \theta )) = ( \frac{\pi}{4} )

Calculate ( v_{0x} ) using ( v_{0x} = v_0 \cdot \cos(\theta) ): [ v_{0x} = \frac{2}{5} \cdot \cos\left(\frac{\pi}{4}\right) ] [ v_{0x} = \frac{2}{5} \cdot \frac{\sqrt{2}}{2} ] [ v_{0x} = \frac{\sqrt{2}}{5} ]

The time taken to reach maximum height (( t )) can be calculated using the formula for vertical motion: [ t = \frac{v_{0y}}{g} ]

where ( v_{0y} ) is the initial velocity in the vertical direction and ( g ) is the acceleration due to gravity (approximately ( 9.8 , \text{m/s}^2 )). Since the projectile reaches its maximum height, its vertical velocity (( v_{0y} )) at that point is 0.

Calculate ( t ) using ( t = \frac{v_{0y}}{g} ): [ v_{0y} = v_0 \cdot \sin(\theta) ] [ v_{0y} = \frac{2}{5} \cdot \sin\left(\frac{\pi}{4}\right) ] [ v_{0y} = \frac{2}{5} \cdot \frac{\sqrt{2}}{2} ] [ v_{0y} = \frac{\sqrt{2}}{5} ]

[ t = \frac{\frac{\sqrt{2}}{5}}{9.8} ] [ t = \frac{\sqrt{2}}{49} ]

Now, use ( d = v_{0x} \cdot t ): [ d = \frac{\sqrt{2}}{5} \cdot \frac{\sqrt{2}}{49} ] [ d = \frac{2}{49} ]

So, the projectile's distance from the starting point when it reaches its maximum height is ( \frac{2}{49} ) 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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