If a projectile is shot at a velocity of #4 m/s# and an angle of #pi/6#, how far will the projectile travel before landing?

2 Answers

Answer:

#1.412\ m#

Explanation:

The horizontal distance/ range #R# traveled by a projectile shot with an initial velocity #u=4\ \text{m/s# at an angle #\theta=\pi/6# with the horizontal is given by following formula

#R=\frac{u^2\sin2\theta}{g}#

#=\frac{4^2\sin(\pi/3)}{9.81}#

#=1.412\ m#

Jul 28, 2018

Answer:

The distance is #=1.41m#

Explanation:

The trajectory of a projectile is given by the equation

#y=xtantheta-(g/(2u^2cos^2theta))x^2#

Here,

The initial velocity is #u=4ms^-1#

The angle is #theta=pi/6#

The acceleration due to gravity is #g=9.8ms^-1#

Therefore,

#y=xtan(pi/6)-(9.8/(2*4^2cos^2(pi/6)))x^2#

#y=0.577x-0.408x^2#

The distance travelled horizontally is when #y=0#

#0=0.577x-0.408x^2#

#x(0.577-0.408x)=0#

#x=0#, this is the initial conditions

#x=0.577/0.408=1.41m#

graph{0.577x-0.408x^2 [-0.18, 2.858, -0.269, 1.249]}