Question

A projectile is projected at an angle of 45°so that it attains maximum height of 25 metre, what is the range of the projectile?

Answer 1

Answer: Range = 100 m.

Explanation:

θ = 45°

h(max) = 25 m

We know,

h(max) = u²sin²θ/2g

=> 25 m = (u² sin² 45°)/(20)

=> 1/2 u² = 500 m²/s²

=> u² = 1000 m²/s² __(a)

Now,

Range = 2u² cosθ sinθ/g

=> R = 2(1000 m²/s² × 1/2)/(10 m/s²)

=> R = 200 × 1/2 m

=> R = 100 m.

More:

• During projectile motion, the x component only keeps changing, i.e., u(x).
• Time of ascent = time of descent = t = u sinθ/g.