Question

A ball is projected at an angle of 45° so as to pass a wall at a distance x1 from the point of

projection. If it falls at a distance x2 on the other side of the wall, find the height of the wall.​

Answer 1

My Take on this

The height of the wall can be related with the Horizontal distance traveled by the equation of the trajectory

y=xtan∆ - gx²(1+tan²∆)/(2u²)

The wall is at distance x1

So thats the horizontal distance we'll use... We need to find its corresponding vertical distance "y"

So

Applying the formula

∆=45°

x=x1

g=9.8ms-²

u=velocity of projection

Substituting

y= x1tan45 - 9.8x1²[1+(tan²45)]/2u²

tan45°=1

y= x1 -4.9x²(1 +1)/u

y= x1 - 4.9x²•2/u²

y= x1 - 9.8x²/u².