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

Adrian Ayala · Accepted Answer

The distance is #=0.459m#

Resolving in the vertical direction #uarr^+#

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

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

#=0.216s#

Resolving in the horizontal direction #rarr^+#

To find the distance, we apply the equation of motion

#s=u_x*t#

#=3cos(1/4pi)*0.216#

#=0.459m#

Parker Abbott · Answer

undefined To find the distance from the starting point when the projectile reaches its maximum height, we first need to determine the time it takes to reach the maximum height. This occurs when the vertical component of the projectile's velocity becomes zero. The time to reach maximum height can be found using the formula:

$$ t = \frac{v_{y0}}{g} $$

Where:
$ v_{y0} $ = initial vertical velocity
$ g $ = acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $)

Given that the projectile is launched at an angle of $ \frac{\pi}{4} $ radians with a speed of $ 3 \, \text{m/s} $, the initial vertical velocity ($ v_{y0} $) can be calculated using trigonometric functions:

$$ v_{y0} = v_0 \sin(\theta) $$

Substitute the values:

$$ v_{y0} = 3 \, \text{m/s} \times \sin\left(\frac{\pi}{4}\right) = 3 \, \text{m/s} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \, \text{m/s} $$

$$ t = \frac{\frac{3\sqrt{2}}{2}}{9.8} \approx 0.306 \, \text{s} $$

Now, to find the horizontal distance traveled at the maximum height, we use the time calculated:

$$ d_{x\text{ max}} = v_{x0} \times t $$

Where:
$ v_{x0} $ = initial horizontal velocity

Given that $ v_{x0} = v_0 \cos(\theta) $:

$$ v_{x0} = 3 \, \text{m/s} \times \cos\left(\frac{\pi}{4}\right) = 3 \, \text{m/s} \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \, \text{m/s} $$

$$ d_{x\text{ max}} = \frac{3\sqrt{2}}{2} \, \text{m/s} \times 0.306 \, \text{s} \approx 0.459 \, \text{m} $$

So, the projectile's distance from the starting point when it reaches its maximum height is approximately $ 0.459 \, \text{m} $.