Recent content by Blox_Nitrates

Sorry, Vox = ucosθ

Voy = usinθ?

Since we are ignoring air resistance than the initial speed v would be the same as initial speed v at max height because the horizontal component's velocity doesn't change. The speed at launch would also be speed V.

The horizontal component would also be V at the initial launch if ignoring air resistance because velocity wouldn't change.

The magnitude of the initial velocity.

There wouldn't be any acceleration right? But then that would leave V=Vo which is true?

V=Vo+at?

It would be moving horizontally for an instant before moving downwards.

Somewhat angled about 60 degrees moving upward, hitting the apex when a=0 and then moving downward towards the ground.

I would see a parabolic function in that case but I still feel like I'm not seeing something that I'm possibly missing.

Sorry. So would the maximal height be somwhere between 45-90 degrees?

I can say that the max height will be vertical and the direction of motion is upwards. Wouldn't that be a 90 degree angle of (pi/2)?

This question throws me off. A projectile's launch speed is five times its speed at maximum height. Find launch angle θo.

I'm not sure if this is correct but I used Savg=TotalDistance/Δt Total Distance = 250m/s(3.03s) = 758m rounded up. The answer does match the one in the book.

45=0*t+.5(9.8)t^2 (45/4.9)^(1/2) = 3.03s Thank you so much for your help LawrenceC! So if I was asked At what horizontal distance from the firing point does it strike the ground? I would use a similar eqn to solve it while plugging in t correct?