If a projectile is shot at a velocity of #15 m/s# and an angle of #pi/4#, how far will the projectile travel before landing?

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Answer 1 · 2016-11-14T19:07:14

The distance is #=22.5m#

#u=15ms^(-1)#

#theta=pi/4#

We use the equation #y(t)=usintheta*t-1/2*g*t^2#

to determine the time of flight.

#0=usintheta*t-1/2g t^2 #

#usintheta=1/2*g*t#

#t=(2usintheta)/g#

The horizontal distance is #d=ucostheta*(2usintheta)/g#

#=(u^2sin2theta)/g#

Let #g=10 ms^(-2)#

#d=15^2*sin(pi/2)/10=225*1/10=22.5 m#

graph{x-(x^2/22.5) [-12.5, 38.8, -0.83, 24.84]}