Question

A projectile crosses two walls of equal height h symmetrically

Answer 1

Since the projectile motion is symmetric,

∴ Total time taken is,

T=2+6=8 s

Also,

T=2vsinθg and

H=v2sin2θ2g

∴H=g84×v2sin2θg2

∴H=g×T28=10×828=80 m Also, for t=2s assuming height of wall is h

h=usinθt−12gt2

∴h=2usinθ−20 .....(i)

and, 2ucosθ=s .....(ii)

for t=6 s

h=+6usinθ−180 .....(iii)

and, 6ucosθ=s+120 ....(iv)

Equating (i) and (iii),

2usinθ−20−6usinθ+180=0

usinθ=1604=40

Put in equation (iii)

h=60 m

This method must be followed to solve this question as it is the easiest way of solving it.

Thus, we can get to the right answer with the correct method easily.

Answer 2

Explanation:

time of flight=8sec

height of each wall=60m

Hmax=80m