A projectile is shot from the ground at an angle of #pi/6 # and a speed of #1 m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

Question

A projectile is shot from the ground at an angle of #pi/6   # and a speed of #1    m/s#. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

William Audy · Accepted Answer

H=0.012m and #d_h=0.044m#

Let the velocity of projection of the object be u with angle of projection #alpha# with the horizontal direction. The vertical component of the velocity of projection is #usinalpha# and the horizontal component is #ucosalpha#

Now if the time of flight be T then the object will return to the ground after T sec and during this T sec its total vertical dosplacement h will be zero. So applying the equation of motion under gravity we can write

#h=usinalphaxxT+1/2gT^2# #=>0=usinalphaxxT-1/2xxgxxT^2# where #g= "acceleration" "dueto"" gravity”# #:.T=(2usinalpha)/g# The horizontal displacement during this T sec is #R=ucosalpha xxT# #R=(u^2sin(2alpha))/g# The time t to reach at the peak is half of time of flight (T) So # t=1/2*T=(usinalpha)/g# The horizontal displacement during time t is #d_h=1/2xx(u^2sin(2alpha))/g# By the problem #u=1m/s;g=9.8m/s^2 and alpha= pi/6# #d_h=1/2xx(1^2sin(2*pi/6)/9.8)=0.044m# If H is the maximum height then # 0^2=u^2sin^2(alpha)-2*g*H# #:. H= (u^2sin^2(alpha))/(2*g)=(1^2sin^2(pi/6))/(2*9.8)=0.012m#

So the distance of the object from the point of projection when it is on the peak Is given by #D=sqrt(d_h^2+H^2)=sqrt((0.044)^2+(0.012^2))=0.045m#

Nicholas Agrusa · Answer

undefined To find the distance from the starting point when the projectile is at its maximum height, you can use the formula for the horizontal range of a projectile:

Range = (initial velocity * time of flight * cos(angle))

At maximum height, the vertical component of velocity is zero, and you can find the time of flight using the formula for the time to reach maximum height:

Time to reach maximum height = (initial vertical velocity) / (gravitational acceleration)

Given:
- Angle = π/6 radians
- Initial velocity = 1 m/s
- Gravitational acceleration = 9.8 m/s²

Calculate the vertical component of the initial velocity:
Initial vertical velocity = initial velocity * sin(angle)

Then, calculate the time to reach maximum height:
Time to reach maximum height = (initial vertical velocity) / (gravitational acceleration)

Finally, calculate the horizontal range:
Range = (initial velocity * time to reach maximum height * cos(angle))

Substitute the given values into the equations and solve for the horizontal range.