A projectile is shot from the ground at an angle of #pi/4 # and a speed of #10 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 #10   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 #=5.7m#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_0=(10)sin(1/4pi)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-(10)sin(1/4pi))^2/(-2g)#

#h_y=(10sin(1/4pi))^2/(2g)=2.55m#

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-(10)sin(1/4pi))/(-9.8)=0.72s#

Resolving in the horizontal direction #rarr^+#

The velocity is constant and #u_x=(10)cos(1/4pi)#

The distance travelled in the horizontal direction is

#s_x=u_x*t=(10)cos(1/4pi)*0.72=5.1m#

The distance from the starting point is

#d=sqrt(h_y^2+s_x^2)=sqrt(2.55^2+5.1^2)=5.7m#

William Audy · Answer

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

$$ \text{Distance} = \text{Initial velocity} \times \text{Time at maximum height} \times \text{Cosine of launch angle} $$

Given:
- Initial velocity ($ v_0 $) = 10 m/s
- Launch angle ($ \theta $) = $ \frac{\pi}{4} $
- Time to reach maximum height ($ t_{\text{max}} $) = $ \frac{v_0 \times \sin(\theta)}{g} $ (where $ g $ is the acceleration due to gravity, approximately $ 9.8 \, \text{m/s}^2 $)

Substituting the given values into the formula:

$$ t_{\text{max}} = \frac{10 \times \sin\left(\frac{\pi}{4}\right)}{9.8} $$
$$ t_{\text{max}} = \frac{10 \times \frac{\sqrt{2}}{2}}{9.8} $$
$$ t_{\text{max}} = \frac{5\sqrt{2}}{9.8} $$

Now, substitute $ t_{\text{max}} $ into the distance formula:

$$ \text{Distance} = 10 \times \frac{5\sqrt{2}}{9.8} \times \cos\left(\frac{\pi}{4}\right) $$

$$ \text{Distance} \approx 5 \, \text{meters} $$

So, the distance from the starting point when the projectile reaches its maximum height is approximately $ 5 \, \text{meters} $.