A projectile is shot at an angle of #pi/8 # and a velocity of # 7 m/s#. How far away will the projectile land?

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Answer 1 · 2018-08-08T12:04:04

The projectile wiil land at #=3.54m#

The trajectory of a projectile is given by the equation

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

Here,

The initial velocity is #u=7ms^-1#

The angle is #theta=pi/8#

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

Therefore,

#y=xtan(pi/8)-(9.8/(2*7^2cos^2(pi/8)))x^2#

#y=0.414x-0.117x^2#

The distance travelled horizontally is when #y=0#

#0=0.414x-0.117x^2#

#x(0.414-0.117x)=0#

#x=0#, this is the initial conditions

#x=0.414/0.117=3.54m#

graph{0.414x-0.117x^2 [-0.18, 2.858, -0.269, 1.249]}