Question

A stone is projected from a point on ground at some angle of projection. At the same time, a bee starts from a point directly above this point of projection at a height 2.5 m horizontally with speed 5√ 2 ms–1. It is found that the stone just touches the bee at ne point. Find range of projectile motion of stone. (a) 10 m (b) 5 m (c) 7.5 m (d) 2.5 m

Answer 1

Given :-

Height = 2.5 m

Velocity = Vx = 5√2 ms-¹

Now,

Velocity in y-direction,

Vy = √2gh

Vy = √2×9.8×2.5

Vy = 7 ms-¹

As we know that,

R = 2Vx•Vy/g

R = 2[5√2 × 7]/10

R = 2(35√2)/10

R = 70√2/10

R = 7√2 m

or

R = 9.8 m

or

R ≈ 10 m

Hence,

Range = R = 10 m

Using Different way,

R = 2u√2h/g

R = 2(5√2)√2×2.5/9.8

R ≈ 10 m

Answer 2

Answer:

Given :-

Height = 2.5 m

Velocity = Vx = 5√2 ms-¹

Now,

Velocity in y-direction,

Vy = √2gh

Vy = √2×9.8×2.5

Vy = 7 ms-¹

As we know that,

R = 2Vx•Vy/g

R = 2[5√2 × 7]/10

R = 2(35√2)/10

R = 70√2/10

R = 7√2 m

or

R = 9.8 m

or

R ≈ 10 m

Hence,

Range = R = 10 m

Using Different way,

R = 2u√2h/g

R = 2(5√2)√2×2.5/9.8

R ≈ 10 m