A projectile is shot from the ground at a velocity of #15 m/s# at an angle of #pi/6#. How long will it take for the projectile to land?

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Answer 1 · 2017-12-17T12:51:24

The time is #=1.53s#

Resolving in the vertical direction #uarr^+#

The initial velocity is #u_0=15sin(1/6pi)ms^-1#

The time to reach the greatest height is #=ts#

Applying the equation of motion

#v=u+at=u- g t #

The time is #t=(v-u)/(-g)=(0-15sin(1/6pi))/(-9.8)=0.765s#

The time taken to land is twice the time to reach the greatest height

#T=2t=2*0.765=1.53s#

Answer 2 · 2017-12-17T12:59:55

#1.531 s#

Here you actually asked about the wandering period of the projectile.

The initial velocity,#v_0=15 ms^(-1)#

The throwing angle of the projectile,θ=#pi/6=30@#
#Suppose,#
The wandering period of the projectile, #color(red)(T)=?#

we know that,
#T=color(blue)((2v_0sintheta)/g)=(2xx15ms^(-1)xxsin30@)/(9.8 ms^(-2))=1.531 s##color(red)((Ans.)#