A projectile is shot at an angle of #pi/8 # and a velocity of # 65 m/s#. How far away will the projectile land?

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Answer 1 · 2018-07-10T06:33:14

#304.539\ m#

The range #R# covered by a projectile shot at an angle #\theta=\pi/8# & a velocity #u=65\ m/s# is given by the following formula

#R=\text{horizontal velocity}\times \text{time of flight}#

#R=(u\cos\theta)({2u\sin\theta}/g)#

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

#=\frac{65^2\sin2(\pi/8)}{9.81}#

#=\frac{4225\cdot 1/\sqrt2}{9.81}#

#=304.539\ m#

Answer 2 · 2018-07-10T08:39:45

The distance is #=304.6m#

Another way of finding the range.

The equation of the trajectory of the projectile is

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

The acceleration due to gravity is #g=9.8ms^-2#

The initial velocity is #u=65ms^-1#

The angle is #theta=pi/8#

Therefore,

#y=xtan(pi/8)-((9.8)/(2*65^5*cos^2(pi/8)))x^2#

#y=0.4142x-0.00136x^2#

The range is when #y=0#

#0.4142x-0.00136x^2=0#

#{(x=0),(x=0.4142/0.00136=304.6m):}#

graph{0.4142x-0.00136x^2 [-91, 390.2, -103.2, 137.5]}