A projectile is shot from the ground at a velocity of #47 m/s# and at an angle of #(pi)/2#. How long will it take for the projectile to land?

2 Answers
Apr 11, 2017

Answer:

The time is #=9.6s#

Explanation:

Resolving in the vertical direction #uarr^+#

initial velocity is #u_y=vsintheta=47*sin(1/2pi)#

Acceleration is #a=-g#

At the maximum height, #v=0#

We apply the equation of motion

#v=u+at#

to calculate the time to reach the greatest height

#0=47sin(1/2pi)-g*t#

#t=47*1/g*sin(1/2pi)#

#=4.8s#

The time for the projectile to land is

#=2*4.8=9.6s#

Apr 11, 2017

Answer:

#9.8s#

Explanation:

If you carefully look into the question:

#pi/2=180/2=90^circ#

#:.# The projectile moves up and comes down vertically

So, To solve, we use

#color(blue)(v=u+at#

Where,

#color(orange)(v="Final velocity"=0#

#color(orange)(u="Initial velocity"=47 m/s#

#color(orange)(a="Acceleration"=-9.8# (As, gravity pulls down)

#color(orange)(t="Time"# (We need to find it)

Let's solve for #t#

#rarrv=u+at#

#rarr0=47-9.8t#

#rarr9.8t=47#

#rarrt=47/9.8#

#color(green)(rArr~~4.8#

Now we have found the time for the projectile to reach its maximum height

So, #"Total time"=4.8*2=9.8s#

Hope this helps.. :)