Question

A man can jump a maximum horizontal distance of 2m, then with what minimum velocity he should
run horizontally along the road in order to cover the same distance in same time.​

Answer 1

Answer:

Given:

A man can jump a max horizontal distance of 2 metres. This implies

" max range " in case of a projectile.

To find:

The velocity with which he should run horizontally to reach same distance in same time.

Concept:

We know that the formula for the range of a projectile =

u²sin(2θ)/g

So for max range , sin(2θ) should have max value = 1.

Hence 2θ = 90°

=> θ = 45°.

so angle of Projection is 45°

Calculation:

We know that horizontal component of velocity in a projectile = u cos(θ)

So max range = 2 metres

=> u²sin(90°)/g = 2

=> u²/g = 2

=> u² = 20

=> u = 2√5 m/s

If the person runs with a velocity equal to the horizontal component of projected velocity, he can reach same distance in same time.

So let the required velocity be "v"

Therefore ,

v = u cos(θ)

=> v = (2√5) × cos(45°)

=> v = (2√5) × (1/√2)

=> v = √10 m/s.

Answer 2

Answer:

Explanation:

A man can jump a maximum horizontal distance 2m, then with what minimum velocity he should run horizontal along the road in order to cover the same distance in same time