A projectile is shot from the ground at an angle of pi/6 π6 and a speed of 1 m/s1ms. When the projectile is at its maximum height, what will its distance, factoring in height and horizontal distance, from the starting point be?

1 Answer
May 12, 2016

H=0.012m and d_h=0.044mdh=0.044m

Explanation:

Let the velocity of projection of the object be u with angle of projection alphaα with the horizontal direction.
The vertical component of the velocity of projection is usinalphausinα and the horizontal component is ucosalphaucosα

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^2h=usinα×T+12gT2
=>0=usinalphaxxT-1/2xxgxxT^20=usinα×T12×g×T2
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