A body is projected from the top of a tower with a velocity baru=3 hat i +4 hat j +5 hat k ms -1, where hat i , hat j and hat k are unit vectors along east, north and vertically upwards respectively. If the height of the tower is 30 m, horizontal range of the body on the ground is (g=10 ms -2)

Question

A body is projected from the top of a tower with a velocity baru=3 hat i +4 hat j +5 hat k ms -1, where hat i , hat j and hat k are unit vectors along east, north and vertically upwards respectively. If the height of the tower is 30 m, horizontal range of the body on the ground is (g=10 ms -2) (A) 15 m (B) 25 m (C) 9 m (D) 12 m

Tardigrade Admin · Accepted Answer

Time in which body is dropped to ground is calculated using following data;

u = 5 m / s

(k is given vertically upward direction)

a = - 10 m / s^{2}

h = - 30 m

Now using the formula

h = u t + \frac{1}{2} a t^{2}

- 30 = 5 t - \frac{1}{2} \times 10 \times t^{2}

\Rightarrow t^{2} - t - 6 = 0

\Rightarrow t = - 2 s

(Not possible)
or

t = 3 s

In this time, projectile moving in East and North with speeds

3 m / s

and

4 m / s

. Distances covered in these directions are;
In east

(x -

coordinate).
- speed

\times time = 3 \times 3 = 9 m

and in North (

y

- coordinate)

= 4 \times 3 = 12 m

So, Projective lando at

(x, y) = (9 m, 12 m)

mark.
So, horizontal range of body on ground is;

Range

= 9^{2} + 1 2^{2}

= 225

= 15 m

Range =√92+122 =√225 =15 m

Q. A body is projected from the top of a tower with a velocity uˉ=3i^+4j^​+5k^ms−1, where i^,j^​ and k^ are unit vectors along east, north and vertically upwards respectively. If the height of the tower is 30m, horizontal range of the body on the ground is (g=10ms−2)

15 m

25 m

9 m

12 m