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?

Answer 1

The distance is #=0.3m#

Resolving in the vertical direction #uarr^+#
The initial velocity is #u_0=9/4sin(1/3pi)ms^-1#

Applying the equation of motion

#v^2=u^2+2as#
At the greatest height, #v=0ms^-1#
The acceleration due to gravity is #a=-g=-9.8ms^-2#

Therefore,

The greatest height is #h_y=s=(0-(9/4sin(1/3pi))^2)/(-2g)#
#h_y=(9/4sin(1/3pi))^2/(2g)=0.194m#
The time to reach the greatest height is #=ts#

Applying the equation of motion

#v=u+at=u- g t #
The time is #t=(v-u)/(-g)=(0-9/4sin(1/3pi))/(-9.8)=0.199s#
Resolving in the horizontal direction #rarr^+#
The velocity is constant and #u_x=9/4cos(1/3pi)#

The distance travelled in the horizontal direction is

#s_x=u_x*t=9/4cos(1/3pi)*0.199=0.224m#

The distance from the starting point is

#d=sqrt(h_y^2+s_x^2)=sqrt(0.194^2+0.224^2)=0.3m#
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Answer 2

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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Answer from HIX Tutor

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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