Vo = 150m/s[85o]
1. Xo = 150*cos85 = 13.07 m/s.
2. Yo = 150*sin85 = 149.4 m/s.
3. X = Xo
4. Y = 0
5. a = g = 9.8 m/s^2.
6. h = ho + (Y^2-Yo^2)/2g
h = 20 + (0-149.4^2)/-19.6 = 1159 m.
Above the plain.
7. Y = Yo + g*t = 0 @ peak.
149.4 -9.8t = 0
9.8t = 149.4
Tr = 15.2 s. = Rise time.
8. h = 0.5g*t^2 = 1159 m.
4.9*t^2 = 1159
t^2 = 236.5
Tf = 15.38 s. = Fall time.
9. T = Tr+Tf = 15.2 + 15.38 = 30.58 s.
10. Y=Yo + g*t = 0 + 9.8*15.38=150.7 m/s
11. V = sqrt(Xo^2+Y^2)
V = sqrt(13.07^2 + 150.7^2) = 151.3 m/s.
12. Range = Xo*(Tr+Tf)
Range = 13.07 * (15.2 + 15.38) = 400 m.
13. Tan A=Y/Xo = 150.7/13.07 = 11.53022
A = 85.0o
A cannon fires a free projectile at an angle 85 from the ground. The cannonball has a speed of 150 m/s and it is launched from a hill 20 m above the plain where it will land
1. What is the x component of the cannonballs initial velocity
2. What is the y component of the cannonballs initial velocity
3. What is the x component of the cannonballs velocity at the peak of its trajectory?
4. What is the y component of the cannonballs velocity at the peak of its trajectory?
5. What is the cannonballs acceleration at the peak of its trajectory?
6. What is the height above the plain of the cannonball at the peak of its trajectory?
7. How long will the cannonball take to reach the peak of its trajectory?
8. How long will the cannonball take to fall from the ground from the peak?
9. What is the total time the cannonball will spend in the air?
10. What will be the y component of the final velocity of the cannonball when it hits the ground?
11. What is the total magnitude of the final velocity of the cannonball when it hits the ground?
12. What is the range of the cannonball?
13. At what angle will the cannonball hit the ground?
1 answer