Projectile Motion Q (1 Viewer)

[Hikari Clover]

A ball is projected from a point on the ground distant a from the foot of a vertical wall of height b. the ball is projected at an elevation A with speed v. if the ball just clears the wall prove that the greatest height is

thx ~~~~

 

[ssglain]

Take notice of the information that you are given. Most of the time, they're there for a reason. One of the conditions that you can establish immediately is the fact that the trajectory passes through P(a, b), which means that at some time T after the projection, x = a & y = b will simultaneously hold. This one fact is the key to solving this question.

Spoiler

x" = 0
x' = VcosA
x = VtcosA

y" = -g
y' = -gt + VsinA
y = -(gt)/2 + VtsinA

The greatest height occurs when y' = 0:
-gt + VsinA = 0 --> t = a/(VcosA)

Put this in the equation for y gives, after simplification:
y(max) = (VsinA)/(2g) ...(1)

Now, we know that at time T:
a = VTcosA ...(2)
b = -(gT)/2 + VTsinA ...(3)

Elimination T from (2) & (3) gives, after simplification:
b = -(ag)/(2VcosA) + atanA --> V = (ag)/[2cosA*(atanA - b)]

Put this in (1) gives, after simplification:
y(max) = (1/4)*(atanA)/(atanA - b) as required.