Questin 1 a cannonball is launched horizontally from the top of an 90.0 m high cliff.
a> how much time will it take for the ball to reach the ground?

b> what is the height of the ball 0.5 sec after of travel?


c. how far away from the base of cliff will the cannonball land?


question 2> a car driver off a horizonal embankment and lands 11.5m from the edge of the embankment in a field that is 3.00m lower than the embankment.,. what is vi aka the velocity of the car when it left the embankment?

ill post question 3 in a min show all work plz


I leave for school in an hour and 15min and had to learn 2 units in an hour. So I need these done b4 then if u want the fg.

This post was edited by Duty on Sep 24 2014 10:44am

If u do these 2 questions i posted ill give u 200fg(100 ea). Offer upped.

This post was edited by Duty on Sep 24 2014 11:00am

Questin 1 a cannonball is launched horizontally from the top of an 90.0 m high cliff.
a> how much time will it take for the ball to reach the ground?
-----------------------------------------
Hitting ground only depends on vertical velocity/acceleration
No initial vertical velocity
Distance= (1/2)(a)(t^2) is usable
90m=(.5)(9.8)(tt)

t=4.29 sec


---------------------------------------
b> what is the height of the ball 0.5 sec after of travel?
same formula can be applied to solve for distance traveled
distance=(.5)(9.8)(.5^2)
1.225 m traveled
90-1.225= 88.78m

c. how far away from the base of cliff will the cannonball land?

It takes ~4.29 seconds for the ball to hit the ground
horizontal distance traveled is simply d=vt
distance= horizontal velocity*4.2857
you did not provide horizontal velocity

question 2> a car driver off a horizonal embankment and lands 11.5m from the edge of the embankment in a field that is 3.00m lower than the embankment.,. what is vi aka the velocity of the car when it left the embankment?


We assume motion is stopped when things land/ hit the ground.

There is no initial vertical velcoity, so Distance= (1/2) (a) (t^2) can be used again

the fall distance is 3m
since we know fall distance and acceleration (9.8 m/s/s) we can solve for time

3=(.5)(9.8)(t^2)

t=.78 seconds
t=.7824607...seconds
for horizontal, distance= horizontal velocity *time can be used

11.5 =v* (.7824607)
horizontal velcoity= 14.697
v=14.70 m/s