Need a Projectile Motion Question Answered please.... toughy... for me! (1 Viewer)

[currysauce]

A coastal defense cannon is positioned at the top of a 150m vertical cliff. It is able o fire a one kilgram shell at a velocity of 120m/s. If fired horizontally out from the cliff, how far out to sea will the shell land.

Second part

The cannon is now aimed at an angle of 45 degrees above horizontal, How far now...


THANKS!

 

[withoutaface]

The initial horizontal velocity is 120m/s
The initial vertical velocity is 0m/s
The overall displacement is -150m
The acceleration due to gravity is -9.8m/s²

For the first part you take the vertical statistics and substitute them into this equation: S=ut+(at²)/2, then maniplulate it to find 't'.

You then substitute 't' and the horizontal velocity and substitute them into the equation range=velocity*time

The second part is essentially the same except the horizontal velocity is 150cos45=150/sqrt2, and the vertical velocity is 150sin45=150/sqrt2