AL_Physics Projectile motion question

Let u be the initial speed

t be the travelling time

Consider horizontal component,

Horizontal velocity, uH = ucos40˚ ms-1

So, range of projectile, sH = utcos40˚ m

Consider the range when the projectile rises to the edge of the wall

Initial vertical velocity, uV = usin40˚ ms-1

By equation of motion,

vv2 = uv2 – 2gs

vv = √(u2sin240˚ - 200) ms-1

Consider the range that crossing the two walls.

Let T be the time required to cross the two walls

So, uTcos40˚ = 20

T = 20 / ucos40˚ s

Once again, by equation of motion,

s’ = vvT – 1/2 gT2

0 = 20 / ucos40˚ √(u2sin240˚ - 200) – 5(20 / ucos40˚)2

Solving, u = 23.6214 ms-1

Now, consider the whole range of the projectile.

Initial vertical velocity, uw = 23.6214sin40˚ = 15.1836 ms-1

By equation of motion,

s = uwt – 1/2 gt2

0 = 15.1836t – 5t2

t = 3.0367 s

So, range of projectile

= 15.1836 X (3.0367)cos40˚

= 55.0 m