How far below an initial straight - line path will a projectile fall in one second?

Question

A) How far below an initial straight - line path will a projectile fall in one second?
B) Does your answer depend on the angle of launch or on the initial speed of the projectile? Defend your answer.


When you answer the question, please also explain WHY, because I don’t just want the answer, I want to learn also =)

Answer 1

You want to learn and understand good for you.

The pull of gravity is constant

distance and object falls = 16.5 x time in seconds to the 2nd power

s = 16.5 t^2

basically it takes an object 1 second to fall about 16.5 feet
if you launch a projectile perfectly horizontally at a velocity of 160 feet per second in 1 second it will go 160 ft horizontally and fall 16.5 feet and hit the ground ( if it's a bowling ball or something it will bounce and roll !)

Answer 2

Let me try:

Things to remember in projectile motion is
1.Velocity has two components, horizontal Vx, and Vertical Vy
2. Horizontal component does not change, remains constant
3. Vertical component changes but has a known acceleration =g (or 9.8 m/s2
To solve this problem, we must first establish initail velocities, and final velocities, and find their angles.

Now step I is the velocity vector, V, V=iVx+jVy, initial firing angle of the projectile is assumed to be O.
Therefore Vx=V CosO, Vy=VSinO, Where O is your angle
Now let us see how velocity changes after One secon.
Vx= same, Vy has changed under acceleration g given by
VyFinal=Vy+1/2gt2, but t=1 second, therefore,
VyFinal=Vy+10/2=Vy+5, but Vy=VsinO,
Hence, Vyfinal=VsinO+5, remeber Vx is the same,
Now Call the new angle O1, TanO1=VyFinal/Vx
or (VsinO+5) / V sinO, OR 1+5/VsinO,
Change in angle would be O-O1,
it is clear that the result is dependent on both the initil velocity
which is V, and initial angle O. Hope I have been of help.
It would be great help had the answer been mentioned too.

Answer 3

On Earth the force of gravity is about 33 feet per second squared or 9.8 meters per second squared. Distance (s) is = to 1/2 acceleration (a) times times squared.

So s= 1/2 (33 feet per second) times 1 squared or about 16.5 feet.