physic help,how can i solve this?

Question

A cannonball,can be launched with an initial velocity of 100 m/s.A target,on the same level as the muzzle of the cannon, is 1000 m away.
Question:
a. can the cannonball reach the target ?
b. at what angle should the cannonball be launched to hit the target?

A car starts from rest and moves along a staright line withan acceleration of a=(3*s^(-1/3))m/s^2,where s is in meters.how can i determine the car's velocuty and position when time is = 10s

many thanks
hope u can help me

Answer 1

Canonball question:

a. The cannonball will not reach the target.
b. Let x = the angle required to hit the target.

100cosx=the horizontal component of the velocity.
Distance=Velocity*Time
Time, t=Distance/Velocity
=1000/100cosx
t =10/cosx

100sinx=the vertical component of the velocity.

v=u+at (where v=0 at the top of the flight of the cannonball, u the the initial vertical component of
the velocity, and a is the equal to -9.8m/s^2.

0=100sinx+(-9.8)t
t=100sinx/9.8 (we multiply this by 2 to get the total time to reach the target). thus
Total t=2*100sinx/9.8

Substitute this value in the equation t=10/cosx:

10/cosx=2*100sinx/9.8
10=2*100sinxcosx/9.8
1=20sinxcosx/9.8
sinxcosx=9.8/20
=0.49
sinx=0.49

(Check your trigonometric table for the nearest value of x, or go to google and search "trigonometric tables."

Car Problem:

Use the formula: v=u+at

Substitute given values:

a=3*s^(-1/3)m/s^2

t=10s sec (I hope you mean this and not 10sec. You have another s in the acceleration, which is in meters. But if you can clear up this matter about the s's there's no problem solving this problem.)

v=0+3*s^(-1/3)(10s)
=3*10s/s^1/3
=30*s^2/3m/s

For the position, calculate the distance S travelled in time 10s sec. (Again the confusing s's!)

S=ut+1/2at^2
=0+1/2*3*s^(-1/3)(10s)^2
=3/2*100*s^2/s^1/3
=150s^5/3meters

Answer 2

a. yes it can
b. 45 degrees

dont know the 2nd question.

Answer 3

Do your own homework, honey.
The answers are in your textbook.

Answer 4

f=ma