1. IF YOU ARE GIVEN A VELOCITY IN THE PROBLEM, IT IS USUALLY vi !!!!!!!!
Projectile Trix
IF YOU ARE GIVEN A VELOCITY IN THE PROBLEM, IT IS USUALLY vi !!!!!!!!
If an object is projected horizontally ……
.... θ = 0°
If an object lands at the same height from which it was projected ……
.... Δdy = 0
For a projectile to travel the maximum horizontal distance …..
…..θ = 45°
thus viy = 0 and vx = vi

2. 1) Given: Δdy = viyt + ½gt2 Δdy = -61 m Δdx = 13 m g = -9.8 m/s2 θ= 0°
-61 = 0t + ½(-9.8)t2
61 m
g = -9.8 m/s2
-61 = -4.9t2
θ= 0°
12.4 = t2
viy = 0 m/s
3.53 s = t
13 m
vx = vi = ? m/s
Δdx = vxt
vx = 3.7 m/s
13 = vx(3.53)

3. viy = vi sinθ = 98 (sin 53°) = 78.3 m/s
2) Given:
vi = 98 m/s
θ= 53°
viy = vi sinθ = 98 (sin 53°) = 78.3 m/s
vx = vi cosθ = 98 (cos 53°) = 59.0 m/s
g = -9.8 m/s2
Δdy = 0 m (lands at the same height from which it was launched)
Δdy = viyt + ½gt2
Δdy = viyt + ½gt2
(max height at half the time)
0 = 78.3t + ½(-9.8)t2
0 = 78.3t – 4.9t2
4.9t2 = 78.3t
+4.9t2
+4.9t2
Δdy max = 78.3(8) + ½(-9.8)(8)2
4.9t
4.9t
Δdy max= 310 m
16 sec = t
Δdx = vxt
Δx = (59.0)(16)
Δdx = 944 m

4. 3) Given: Δdy = viyt + ½gt2 Δdy = -1.225 m Δdx = 0.40 m g = -9.8 m/s2
-1.225= 0t + ½(-9.8)t2
g = -9.8 m/s2
= -4.9t2
θ= 0°
0.25 = t2
viy = 0 m/s
0.50 s = t
vx = vi = ? m/s
Δdx = vxt
vx = 0.80 m/s
0.40 = vx(0.50)

5. 4) Given: Δdy = viyt + ½gt2 Δdy = -176.4 m Δdx = vxt vi= 22 m/s
-176.4= 0t + ½(-9.8)t2
Δdx = (22)(6)
g = -9.8 m/s2
= -4.9t2
θ= 0°
36 = t2
Δdx = 132 m
viy = 22(sin 0) = 0 m/s
6 s = t
vx = 22 (cos 0) = 22 m/s
Δdy = viyt + ½gt2
5) Given:
viy= 350 (sin0) = 0 m/s
-1500= 0t + ½(-9.8)t2
θ= 0°
vx = 350 (cos 0 ) = 350 m/s
vi= 350 m/s
-1500 = -4.9t2
Δdy = m
17.5 s = t
306 = t2
g = -9.8 m/s2

6. 6)
Horizontal speed is constant everywhere
Vertical speed is largest at the lowest point and smallest (zero) at the top
Projectiles only accelerate in the vertical direction, so the acceleration is constant and down everywhere (gravity)

7. viy = vi sinθ = 50 (sin 45°) = 35.4 m/s
7) Given:
vi = 50 m/s
θ= 45°
viy = vi sinθ = 50 (sin 45°) = 35.4 m/s
vx = vi cosθ = 50 (cos 45°) = 35.4 m/s
g = -9.8 m/s2
Δdy = 0 m (identical landing ramp)
Δdy = viyt + ½gt2
0 = 35.4t – 4.9t2
0 = 35.4t + ½(-9.8)t2
+4.9t2
+4.9t2
4.9t2 = 35.4t
4.9t
4.9t
7.2 sec = t
Δdx = vxt
Δdx = (35.4)(7.2)
Δdx = 255 m

8. viy = vi sinθ = 20 (sin 40°) = 12.9 m/s
8) Given:
vi = 20 m/s
θ= 40°
20 m
viy = vi sinθ = 20 (sin 40°) = 12.9 m/s
vx = vi cosθ = 20 (cos 40°) = 15.3 m/s
g = -9.8 m/s2
Δdx = vxt
20 = (15.3)(t)
Δdx = 20 m
Δdy = viyt + ½gt2
1.3 sec = t
Δdy = 8.5 m
Δdy = 12.9(1.3) + ½(-9.8)(1.3)2
9) Given:
Δdy = viyt + ½gt2
θ= 0°
viy= 0 cm/s
vx = vi = 20 cm/s
-120= 0t + ½(-980)t2
Δdy = -120 cm
0.5 s = t
Δx = 10 cm
Δdx = vxt
g = -980 cm/s2
Δdx = (20)(0.5)