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?

Question

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?

Isabella Alvarez · Accepted Answer

The distance is #=0.13m#

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=3/2*sin(1/4pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=3/2sin(1/4pi)-g*t#

#t=3/(2g)*sin(1/4pi)#

#=0.108s#

The greatest height is

#h=(3/2sin(1/4pi))^2/(2g)=0.057m#

Resolving in the horizontal direction #rarr^+#

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

#s=u_x*t#

#=3/2cos(1/4pi)*0.108#

#=0.115m#

The distance from the starting point is

#d=sqrt(h^2+s^2)#

#=sqrt(0.057^2+0.115^2)#

#=0.13m#

Nathan Ashcroft · Answer

undefined 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.