Question

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?

Answer 1

(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.

Answer 2

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,

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)