A particle is projected at an angle of 45 degree from 8 m before the foot of a wall just touches the top of the wall and falls on the ground on the opposite side at a distance 4 m from it. The height of the wall is:-
a. 2/3m b. 4/3m c. 8/3m d. 3/4m

Imagine throwing a ball at 45° toward a wall that stands 8 m away, and watching it just skim the top before landing another 4 m beyond. What height must that wall have? Let’s peel back each layer.

Setting Up the Path

Because the launch angle is 45°, the total horizontal distance from launch to landing is governed by the familiar range formula for projectile motion:

• Range R = (v²·sin2θ)/g, and with θ = 45°, sin2θ = 1, so R = v²/g.
• Here R = 8 m + 4 m = 12 m, so v² = 12 g.

Describing the Curve

Every point (x, y) on the trajectory satisfies:

With θ = 45°, tanθ = 1 and cos²θ = ½, that simplifies neatly to:

Finding the Wall’s Height

At the wall’s foot, x = 8 m, so the height y is:

Answer Choice

c) 8/3 m