A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
What variable are you looking for?
Δx
Δy
Vf
Vi
t
a
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ In which direction is the acceleration?
30o
x-direction
y-direction
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ What is the y-direction acceleration? (ay = ?)
30 m/s
0 m/s2
-9.8 m/s2
45 m/s
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) In which direction is the velocity?
30o
x-direction
y-direction
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi = 45 m/s at 30o Since the velocity and acceleration are not in the same direction, what do you do next?
Use the kinematic equations at 30o.
Resolve the velocity into its x and y components.
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi = 45 m/s at 30o
Resolve the velocity into its x and y components.
Vx = _________
Vx = Vi(cosθ)
Vx = Vi(sinθ)
Vx = Vi(tanθ)
Vx = - 9.8 m/s2
Vx = 45 m/s
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi = 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s
Vy = _________
Vy = Vi(cosθ)
Vy = Vi(sinθ)
Vy = Vi(tanθ)
Vy = - 9.8 m/s2
Vy = 45 m/s
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s Vy = Vi(sinθ) = 22.5 m/s
Since you are solving for Δx, what is the acceleration in the x-direction?
30 m/s
0 m/s2
-9.8 m/s2
45 m/s
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s Vy = Vi(sinθ) = 22.5 m/s ax = 0 m/s2 What equation will you use to find Δx?
vf = vi + aΔt
Δx = viΔt + (½)a(Δt)2
vf2 = vi2 + 2aΔx
Δx = (½)(vi + vf)Δt
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s Vy = Vi(sinθ) = 22.5 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Unfortunately we don't know "Δt", so find it in the y-direction. We need to complete our y-direction variables. (And keep our directions separated) t = _______ What is Δy?
45 m
30 m
0 m
22.5 m
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ ay = -9.8 m/s2 (vertical) vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s Vy = Vi(sinθ) = 22.5 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Unfortunately we don't know "Δt", so find it in the y-direction. We need to complete our y-direction variables. (And keep our directions separated) y-direction t = _______ Δy = 0 m What is Vi in the y-direction?
45 m/s
22.5 m/s
39 m/s
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Find "t" in the y-direction t = _______ Δy = 0 m Vyi = 22.5 m/s ay = -9.8 m/s2 What is the final velocity in the y-direction?
45 m/s
39 m/s
22.5 m/s
-39 m/s
- 22.5 m/s
-45 m/s
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Find "t" in the y-direction t = _______ Δy = 0 m Vyi = 22.5 m/s ay = -9.8 m/s2 Vyf = -22.5 m/s What equation will you use to find "t"?
vf = vi + aΔt
Δx = viΔt + (½)a(Δt)2
vf2 = vi2 + 2aΔx
Δx = (½)(vi + vf)Δt
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Find "t" in the y-direction t = _______ Δy = 0 m Vyi = 22.5 m/s ay = -9.8 m/s2 Vyf = -22.5 m/s vf = vi + aΔt Which of the following is correct?
-22.5 = 22.5 - 9.8t
22.5 = -22.5 - 9.8t
-22.5 = 39 - 9.8t
45 = 22. 5 - 9.8t
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Find "t" in the y-direction t = _______ Δy = 0 m Vyi = 22.5 m/s ay = -9.8 m/s2 Vyf = -22.5 m/s vf = vi + aΔt -22.5 = 22.5 - 9.8t Solve for t.
t = 4.6 sec
t = 1.8 sec
t = 0 sec
t = 1.39 sec
A cannonball is shot at 30o going 45 m/s. If it is shot from the ground and lands on the ground, find its range.
Δx = _____ vi= 45 m/s at 30o Vx = Vi(cosθ) = 39 m/s ax = 0 m/s2 What equation will you use to find Δx? Δx = Vxt
Find "t" in the y-direction Δy = 0 m Vyi = 22.5 m/s ay = -9.8 m/s2 Vyf = -22.5 m/s vf = vi + aΔt -22.5 = 22.5 - 9.8t t = 4.6 sec So now go back and solve for Δx.