A stone is projected from the top tower at an angle of 30 degrees to the horizontal with a velocity of 40m/s. It reaches the ground in 5 seconds. What is the height of the tower and the range of the stone using?

(b)

The horizontal component of velocity is constant so we can write:

#sf(d=v_(x)t)#

#sf(d=vcosthetat)#

#sf(d=40xxcos30xx5=173color(white)(x)m)#

(a)

I'll set the point of launch to be the origin and use the convention that "up is +ve".

We can use:

#sf(s=ut+1/2at^2)#

This becomes:

#sf(h=v_(y)t-1/2"g"t^2)#

#sf(h=vsinthetat-1/2"g"t^2)#

#sf(h=40sin30xx5-1/2xx9.81xx5^2color(white)(x)m)#

#sf(h=100-122.625=-22.625color(white)(x)m)#

This means that the top of the tower is 23 m above the ground.

Well,let's see the pathway of the stone i.e how it reached the ground,

So,it followed a projectile motion from #A# to #B# and then after reaching the same height (#B#) to that of its point of projection,it not only fell down but also proceeded forwards o reach #D#,the forward moveemnt was because of its constant horizontal component of velocity.

So,total time taken to reach the ground is = the time taken to complete the projectile motion + time taken to cover along the path #BD#

So,total time for projectile motion is given as, #t=(2u sin theta)/g#

where, #u# is the velocity of projection and #theta# is the angle of projection.

so putting #u=40# and #theta=30# we get,

#t=4.1 s#

So,rest #(5-4.1)=0.9s# was taken for the stone to move the rest of the path.

Now,when the projectile reaches point #B# it has the same vertical component of velocity downwards(#u sin 30#) that it had upwards during projection.

so,if the height of the tower is #h# then we can say,

#h= 40 sin 30 t + 1/2 g t^2# (using, #h=ut+1/2g t^2#)

putting, #t=0.9# we get, #h=21.97m#

So,in this total #5s# time it has moved along #ED#,which is its horizontal range (#R#),so we can write, #R=u cos theta *5=173.2m# (using #s=vt#,as horizontally its velocity was constant)