Should be the second answer (b)
Answer:
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.
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