Physics problems. Can you please help?

Question

1) A projectile leaves the top of an 80-meter-high building horizantally with a speed of 40 m/s (a) How far horizantally will it go before hitting the ground? (b) What is the final horizantal velocity (hfx) before hitting the ground?

A small rocket is shot at a 45 degree angle from the ground at 4.5 m/s. What are its x and y components

please include the equation(s)/idea and process you used to figure it out. serious answers only!

sorry, didn't make it clear that the problem said to ignore air resistance. thanks to theyuks for pointing that out

Answer 1

Okay, the height of an object in free fall at any time t is given by:

h(t) = -9.8 t^2 + vt + p, where h = height, v=initial vertical velocity, and p = initial height. So if your projectile is travelling horizontally at time t=0, it has no initial vertical velocity, and the equation becomes h(t) = 0 = -9.8 t^2 + 80. Solve for t and you get 2.857 seconds.

Since you don't mention the shape of the projectile, I can't really tell you how wind resistance would affect it, so I'll have to assume for the sake of this problem that it is not a factor. Therefore, the projectile's forward momentum does not slow at all, and it continues at 40 m/s outwards as it also accelerates down. so 40 m/s X 2.857 seconds = 114.29 m of travel before hitting the ground.

As for your rocket, just draw its trajectory out. A 45 degree angle of incline means your rocket forms an isoceles triangle with the ground, with it's vector being the hypotenuse. Therefore, each of the sides (the x and y components) is 0.707 times the hypotenuse (since sin and cos of 45 degrees is (sqrt 2)/2), or 3.182 m/s.

Answer 2

solve as though it was in freefall, determine the amount of time it takes to hit the ground from freefall and multiply that by 40. Like
-4.9t^2+80=0 and solve for t. Then d=40*t
The second one is x=4.5*cos(45) and y=4.5*sin(45) which should equal each other in this case
EDIT: @theyuks you forgot the 1/2 in 1/2*a*t^2

Answer 3

good