Question

A bomber plane moving at a horizontal speed of 500 m/s and a bomb released from it, strikes the ground in 10 sec. Angle at which it strikes the ground will be?

Answer 1

Given :-

Horizontal Speed of plane = Vx = 500 m/s

Now, we have to find vertical speed of the plane.

Vy = gt

Vy = 10 × 10

Vy = 100 m/s

Now for angle of Projection.

tanØ = Vy/Vx

tanØ = 100/500

tanØ = 1/5

Ø = tan-¹(1/5)

Hence,

The angle of Projection = Ø = tan-¹(1/5).

Answer 2

Answer:

Angle = tan ⁻¹ (1/5 )

Explanation:

Considering that you know the concept of v resultant for vx and vy. Now, we know that, horizontal velocity remains constant at all places in the horizontal direction ( Vx = constant = 500 m/s ). This value can be variable only for vertical velocity.

Since,             Vy = gt

Hence,           Vy = 10 × 10    ( Given, t= 10 s, & g = 10 m/s² )

Now, let the angle at which the bomb strikes the ground be ∅.

So,

                  tan ∅ = Vy/ Vx

                  tan ∅ = 10 × 10 / 500

                  tan ∅ = 100/500

                         ∅ = tan ⁻¹ 1/5

Therefore, angle at which it strikes the ground will be tan ⁻¹ 1/5.