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?

2 Answers
Mar 3, 2018

(a)

23 m

(b)

173 m

Explanation:

(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.

Mar 3, 2018

height of the tower = 21.97m

Range of motion=173.2 m

Explanation:

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

enter image source here

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)