What Angle Should a Projectile Be Fired to Clear a 25m Wall?

[silly.kid]

Homework Statement

Ok so i need to work out at what angle a projectile is fired at to clear a wall that is 25m away and 15m tall. the projectile is traveling at 53m/s

Homework Equations

x=(Vx)(t)

y=(Vy)(t)-1/2gt^2

t=x/Vx

The Attempt at a Solution

so i started by substituing in t

y= [(x)(Vy)]/-1/2g(x^2/Vx^2)

after a lot of guessing and checking i worked out the 2 angles that it could be were 33.5351 and 87.4287

so I am already ok as far as the answers go, i was just wondering how the equation i used can be derived to find θ by substituing in Vcosθ and Vsinθ. this is beyond me. i would just like to know for future reference

thanks

 

[Fewmet]

Do you see how to find the vertical and horizontal components on the 53 m/s velocity?

I haven't worked all the way through this, but think about what happens if you assume the projectile goes just high enough to clear the wall. That means the final vertical velocity is 0 m/s when the final position is 15 m.

Can you find the time it reaches 15 m?