# A particle is projected at an angle of 45 from a point lying 2m from the foot of a wall. It just touches the top of the wall and falls on the ground 4m from it. What is the height of the wall?

1
by shivam1

Log in to add a comment

by shivam1

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

Projectile Range= R = 2 m + 4m = 6 m

Let g = 10 m/sec²

let θ = angle of projection = 45 deg

x = u cosθ t

y = u sin θ t - 1/2 g t²

=> y = tan θ x - (g x²) /(2 u² cos² θ) --- trajectory of the particle

y = x - 10 x²/u²

we know that at x = Range = 6m, y = 0

=> 0 = 6 - 10 * 36/u²

=> u² = 60 units

equation of projectile = y = x - x²/6

we know that the wall is at x = 2 meters from the point of projection.

* y = height of wall = 2 - 10 * 2² /60 = 1.33 meters *

Let g = 10 m/sec²

let θ = angle of projection = 45 deg

x = u cosθ t

y = u sin θ t - 1/2 g t²

=> y = tan θ x - (g x²) /(2 u² cos² θ) --- trajectory of the particle

y = x - 10 x²/u²

we know that at x = Range = 6m, y = 0

=> 0 = 6 - 10 * 36/u²

=> u² = 60 units

equation of projectile = y = x - x²/6

we know that the wall is at x = 2 meters from the point of projection.